{{task|Arithmetic operations}} Compute the '''n'''^th term of a [[wp:Series (mathematics)|series]], i.e. the sum of the '''n''' first terms of the corresponding [[wp:sequence|sequence]].

Informally this value, or its limit when '''n''' tends to infinity, is also called the ''sum of the series'', thus the title of this task.

For this task, use: :::::: $S\_n\; =\; \backslash sum\_\{k=1\}^n\; \backslash frac\{1\}\{k^2\}$

:: and compute $S\_\{1000\}$

This approximates the [[wp:Riemann zeta function|zeta function]] for S=2, whose exact value

:::::: $\backslash zeta(2)\; =\; \{\backslash pi^2\backslash over\; 6\}$

is the solution of the [[wp:Basel problem|Basel problem]].

360 Assembly

*        Sum of a series           30/03/2017
SUMSER   CSECT
         USING  SUMSER,12          base register
         LR     12,15              set addressability
         LR     10,14              save r14
         LE     4,=E'0'            s=0
         LE     2,=E'1'            i=1
       DO WHILE=(CE,2,LE,=E'1000') do i=1 to 1000
         LER    0,2                  i
         MER    0,2                  *i
         LE     6,=E'1'              1
         DER    6,0                  1/i**2
         AER    4,6                  s=s+1/i**2
         AE     2,=E'1'              i=i+1
       ENDDO    ,                  enddo i
         LA     0,4                format F13.4
         LER    0,4                s
         BAL    14,FORMATF         call formatf
         MVC    PG(13),0(1)        retrieve result
         XPRNT  PG,80              print buffer
         BR     10                 exit
         COPY   FORMATF            formatf code
PG       DC     CL80' '            buffer
         END    SUMSER

{{out}}

       1.6439

ACL2

(defun sum-x^-2 (max-x)
   (if (zp max-x)
       0
       (+ (/ (* max-x max-x))
          (sum-x^-2 (1- max-x)))))

ActionScript

function partialSum(n:uint):Number
{
	var sum:Number = 0;
	for(var i:uint = 1; i <= n; i++)
		sum += 1/(i*i);
	return sum;
}
trace(partialSum(1000));

Ada

with Ada.Text_Io; use Ada.Text_Io;

procedure Sum_Series is
   function F(X : Long_Float) return Long_Float is
   begin
      return 1.0 / X**2;
   end F;
   package Lf_Io is new Ada.Text_Io.Float_Io(Long_Float);
   use Lf_Io;
   Sum : Long_Float := 0.0;
   subtype Param_Range is Integer range 1..1000;
begin
   for I in Param_Range loop
      Sum := Sum + F(Long_Float(I));
   end loop;
   Put("Sum of F(x) from" & Integer'Image(Param_Range'First) &
      " to" & Integer'Image(Param_Range'Last) & " is ");
   Put(Item => Sum, Aft => 10, Exp => 0);
   New_Line;
end Sum_Series;

Aime

real
Invsqr(real n)
{
    1 / (n * n);
}

integer
main(void)
{
    integer i;
    real sum;

    sum = 0;

    i = 1;
    while (i < 1000) {
        sum += Invsqr(i);
        i += 1;
    }

    o_real(14, sum);
    o_byte('\n');

    0;
}

ALGOL 68

{{works with|ALGOL 68|Revision 1 - no extensions to language used}}

{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}}

{{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d]}}

MODE RANGE = STRUCT(INT lwb, upb);

PROC sum = (PROC (INT)LONG REAL f, RANGE range)LONG REAL:(
  LONG REAL sum := LENG 0.0;
  FOR i FROM lwb OF range TO upb OF range DO
     sum := sum + f(i)
  OD;
  sum
);

test:(
  RANGE range = (1,100);
  PROC f = (INT x)LONG REAL: LENG REAL(1) / LENG REAL(x)**2;
  print(("Sum of f(x) from", lwb OF range, " to ",upb OF range," is ", SHORTEN sum(f,range),".", new line))
)

Output:

Sum of f(x) from         +1 to        +100 is +1.63498390018489e  +0.

APL

      +/÷2*⍨⍳1000
1.64393

AppleScript

{{Trans|JavaScript}} {{Trans|Haskell}}

-- SUM OF SERIES ------------------------------------------

-- seriesSum :: Num a => (a -> a) -> [a] -> a
on seriesSum(f, xs)
    script go
        property mf : |λ| of mReturn(f)
        on |λ|(a, x)
            a + mf(x)
        end |λ|
    end script

    foldl(go, 0, xs)
end seriesSum


-- TEST ---------------------------------------------------

-- inverseSquare :: Num -> Num
on inverseSquare(x)
    1 / (x ^ 2)
end inverseSquare

on run
    seriesSum(inverseSquare, enumFromTo(1, 1000))

    --> 1.643934566682
end run


-- GENERIC FUNCTIONS --------------------------------------

-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
    if m ≤ n then
        set lst to {}
        repeat with i from m to n
            set end of lst to i
        end repeat
        lst
    else
        {}
    end if
end enumFromTo

-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
    tell mReturn(f)
        set v to startValue
        set lng to length of xs
        repeat with i from 1 to lng
            set v to |λ|(v, item i of xs, i, xs)
        end repeat
        return v
    end tell
end foldl

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
    if class of f is script then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn

{{Out}}

## AutoHotkey

AutoHotkey allows the precision of floating point numbers generated by math operations to be adjusted via the SetFormat command. The default is 6 decimal places.

```autohotkey
SetFormat, FloatFast, 0.15
While A_Index <= 1000
 sum += 1/A_Index**2
MsgBox,% sum  ;1.643934566681554

AWK

$ awk 'BEGIN{for(i=1;i<=1000;i++)s+=1/(i*i);print s}'
1.64393

BASIC

{{works with|QuickBasic|4.5}}

function s(x%)
   s = 1 / x ^ 2
end function

function sum(low%, high%)
   ret = 0
   for i = low to high
      ret = ret + s(i)
   next i
   sum = ret
end function
print sum(1, 1000)

BBC BASIC

      FOR i% = 1 TO 1000
        sum += 1/i%^2
      NEXT
      PRINT sum

bc

define f(x) {
    return(1 / (x * x))
}

define s(n) {
    auto i, s

    for (i = 1; i <= n; i++) {
        s += f(i)
    }

    return(s)
}

scale = 20
s(1000)

{{Out}}

1.64393456668155979824

Befunge

Emulates fixed point arithmetic with a 32 bit integer so the result is not very accurate.

05558***>::"~"%00p"~"/10p"( }}2"*v
v*8555$_^#!:-1+*"~"g01g00+/*:\***<
<@$_,#!>#:<+*<v+*86%+55:p00<6\0/**
   "."\55+%68^>\55+/00g1-:#^_$

{{out}}

1.643934

Bracmat

( 0:?i
& 0:?S
& whl'(1+!i:~>1000:?i&!i^-2+!S:?S)
& out$!S
& out$(flt$(!S,10))
);

Output:

8354593848314...../5082072010432.....  (1732 digits and a slash)
1,6439345667*10E0

Brat

p 1.to(1000).reduce 0 { sum, x | sum + 1.0 / x ^ 2 }  #Prints 1.6439345666816

C

#include <stdio.h>

double Invsqr(double n)
{
	return 1 / (n*n);
}

int main (int argc, char *argv[])
{
	int i, start = 1, end = 1000;
	double sum = 0.0;

	for( i = start; i <= end; i++)
		sum += Invsqr((double)i);

	printf("%16.14f\n", sum);

	return 0;
}

C++

#include <iostream>

double f(double x);

int main()
{
    unsigned int start = 1;
    unsigned int end = 1000;
    double sum = 0;

    for( unsigned int x = start; x <= end; ++x )
    {
        sum += f(x);
    }
    std::cout << "Sum of f(x) from " << start << " to " << end << " is " << sum << std::endl;
    return 0;
}


double f(double x)
{
    return ( 1.0 / ( x * x ) );
}

C#

class Program
{
    static void Main(string[] args)
    {
        // Create and fill a list of number 1 to 1000

        List<double> myList = new List<double>();
        for (double i = 1; i < 1001; i++)
        {
            myList.Add(i);
        }
        // Calculate the sum of 1/x^2

        var sum = myList.Sum(x => 1/(x*x));

        Console.WriteLine(sum);
        Console.ReadLine();
    }
}

An alternative approach using Enumerable.Range() to generate the numbers.

class Program
{
    static void Main(string[] args)
    {
        double sum = Enumerable.Range(1, 1000).Sum(x => 1.0 / (x * x));

        Console.WriteLine(sum);
        Console.ReadLine();
    }
}

CLIPS

(deffunction S (?x) (/ 1 (* ?x ?x)))
(deffunction partial-sum-S
  (?start ?stop)
  (bind ?sum 0)
  (loop-for-count (?i ?start ?stop) do
    (bind ?sum (+ ?sum (S ?i)))
  )
  (return ?sum)
)

Usage:

CLIPS> (partial-sum-S 1 1000)
1.64393456668156

Clojure

(reduce + (map #(/ 1.0 % %) (range 1 1001)))

COBOL

       IDENTIFICATION DIVISION.
       PROGRAM-ID. sum-of-series.

       DATA DIVISION.
       WORKING-STORAGE SECTION.
       78  N                       VALUE 1000.

       01  series-term             USAGE FLOAT-LONG.
       01  i                       PIC 9(4).

       PROCEDURE DIVISION.
           PERFORM VARYING i FROM 1 BY 1 UNTIL N < i
               COMPUTE series-term = series-term + (1 / i ** 2)
           END-PERFORM

           DISPLAY series-term

           GOBACK
           .

{{out}}

1.643933784000000120

CoffeeScript

console.log [1..1000].reduce((acc, x) -> acc + (1.0 / (x*x)))

Common Lisp

(loop for x from 1 to 1000 summing (expt x -2))

Crystal

{{trans|Ruby}}

puts (1..1000).sum{ |x| 1.0 / x ** 2 }
puts (1..5000).sum{ |x| 1.0 / x ** 2 }
puts (1..9999).sum{ |x| 1.0 / x ** 2 }
puts Math::PI ** 2 / 6

{{out}}

1.6439345666815615
1.6447340868469014
1.6448340618480652
1.6449340668482264

D

More Procedural Style

import std.stdio, std.traits;

ReturnType!TF series(TF)(TF func, int end, int start=1)
pure nothrow @safe @nogc {
    typeof(return) sum = 0;
    foreach (immutable i; start .. end + 1)
        sum += func(i);
    return sum;
}

void main() {
    writeln("Sum: ", series((in int n) => 1.0L / (n ^^ 2), 1_000));
}

{{out}}

Sum: 1.64393

More functional Style

Same output.

import std.stdio, std.algorithm, std.range;

enum series(alias F) = (in int end, in int start=1)
    pure nothrow @nogc => iota(start, end + 1).map!F.sum;

void main() {
    writeln("Sum: ", series!q{1.0L / (a ^^ 2)}(1_000));
}

Dart

{{trans|Scala}}

main() {
  var list = new List<int>.generate(1000, (i) => i + 1);

  num sum = 0;

  (list.map((x) => 1.0 / (x * x))).forEach((num e) {
    sum += e;
  });
  print(sum);
}

{{trans|F#}}

f(double x) {
  if (x == 0)
    return x;
  else
    return (1.0 / (x * x)) + f(x - 1.0);
}

main() {
  print(f(1000));
}

Delphi

unit Form_SumOfASeries_Unit;

interface

uses
  Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
  Dialogs, StdCtrls;

type
  TFormSumOfASeries = class(TForm)
    M_Log: TMemo;
    B_Calc: TButton;
    procedure B_CalcClick(Sender: TObject);
  private
    { Private-Deklarationen }
  public
    { Public-Deklarationen }
  end;

var
  FormSumOfASeries: TFormSumOfASeries;

implementation

{$R *.dfm}

function Sum_Of_A_Series(_from,_to:int64):extended;
begin
  result:=0;
  while _from<=_to do
  begin
    result:=result+1.0/(_from*_from);
    inc(_from);
  end;
end;

procedure TFormSumOfASeries.B_CalcClick(Sender: TObject);
begin
  try
    M_Log.Lines.Add(FloatToStr(Sum_Of_A_Series(1, 1000)));
  except
    M_Log.Lines.Add('Error');
  end;
end;

end.

{{out}}

1.64393456668156

DWScript

var s : Float;
for var i := 1 to 1000 do
   s += 1 / Sqr(i);

PrintLn(s);

E

pragma.enable("accumulator")
accum 0 for x in 1..1000 { _ + 1 / x ** 2 }

EchoLisp

(lib 'math) ;; for (sigma f(n) nfrom nto) function
(Σ (λ(n) (// (* n n))) 1 1000)
;; or
(sigma (lambda(n) (// (* n n))) 1 1000)
    → 1.6439345666815615

(// (* PI PI) 6)
    → 1.6449340668482264

Eiffel

note
	description: "Compute the n-th term of a series"

class
	SUM_OF_SERIES_EXAMPLE

inherit
	MATH_CONST

create
	make

feature -- Initialization

	make
		local
			approximated, known: REAL_64
		do
			known := Pi^2 / 6

			approximated := sum_until (agent g, 1001)
			print ("%Nzeta function exact value: %N")
			print (known)
			print ("%Nzeta function approximated value: %N")
			print (approximated)
		end

feature -- Access

	g (k: INTEGER): REAL_64
			-- 'k'-th term of the serie
		require
			k_positive: k > 0
		do
			Result := 1 / (k * k)
		end

	sum_until (s: FUNCTION [ANY, TUPLE [INTEGER], REAL_64]; n: INTEGER): REAL_64
			-- sum of the 'n' first terms of 's'
		require
			n_positive: n > 0
			one_parameter: s.open_count = 1
		do
			Result := 0
			across 1 |..| n as it loop
				Result := Result + s.item ([it.item])
			end
		end

end

Elixir

iex(1)> Enum.reduce(1..1000, 0, fn x,sum -> sum + 1/(x*x) end)
1.6439345666815615

Elena

ELENA 4.x :

import system'routines;
import extensions;

public program()
{
    var sum := new Range(1, 1000).selectBy:(x => 1.0r / (x * x)).summarize(new Real());

    console.printLine:sum
}

{{out}}

1.643933566682

Emacs Lisp

(defun serie (n)
  (if (< 0 n)
      (apply '+ (mapcar (lambda (k) (/ 1.0 (* k k) )) (number-sequence 1 n) ))
    (error "input error") ))

(insert (format "%.10f" (serie 1000) ))

Output:

1.6439345667

Erlang

lists:sum([1/math:pow(X,2) || X <- lists:seq(1,1000)]).

Euphoria

{{works with|Euphoria|4.0.0}} This is based on the [[BASIC]] example.

function s( atom x )
	return 1 / power( x, 2 )
end function

function sum( atom low, atom high )
	atom ret = 0.0
	for i = low to high do
		ret = ret + s( i )
	end for
	return ret
end function

printf( 1, "%.15f\n", sum( 1, 1000 ) )

Ezhil

## இந்த நிரல் தொடர் கூட்டல் (Sum Of Series) என்ற வகையைச் சேர்ந்தது

## இந்த நிரல் ஒன்று முதல் தரப்பட்ட எண் வரை 1/(எண் * எண்) எனக் கணக்கிட்டுக் கூட்டி விடை தரும்

நிரல்பாகம் தொடர்க்கூட்டல்(எண்1)

  எண்2 = 0

  @(எண்3 = 1, எண்3 <= எண்1, எண்3 = எண்3 + 1) ஆக

    ## ஒவ்வோர் எண்ணின் வர்க்கத்தைக் கணக்கிட்டு, ஒன்றை அதனால் வகுத்துக் கூட்டுகிறோம்

    எண்2 = எண்2 + (1 / (எண்3 * எண்3))

  முடி

  பின்கொடு (எண்2)

முடி

அ = int(உள்ளீடு("ஓர் எண்ணைச் சொல்லுங்கள்: "))

பதிப்பி "நீங்கள் தந்த எண் " அ
பதிப்பி "அதன் தொடர்க் கூட்டல் " தொடர்க்கூட்டல்(அ)

Factor

1000 [1,b] [ >float sq recip ] map-sum

Fantom

Within 'fansh':

fansh> (1..1000).toList.reduce(0.0f) |Obj a, Int v -> Obj| { (Float)a + (1.0f/(v*v)) }
1.6439345666815615

Fish

0&aaa**>::*1$,&v
    ;n&^?:-1&+ <

=={{header|Fōrmulæ}}==

In [http://wiki.formulae.org/Sum_of_a_series this] page you can see the solution of this task.

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions more info]). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.

The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.

Forth

: sum ( fn start count -- fsum )
  0e
  bounds do
    i s>d d>f dup execute f+
  loop drop ;

:noname ( x -- 1/x^2 ) fdup f* 1/f ;   ( xt )
1 1000 sum f.       \ 1.64393456668156
pi pi f* 6e f/ f.   \ 1.64493406684823

Fortran

In ISO Fortran 90 and later, use SUM intrinsic:

real, dimension(1000) :: a = (/ (1.0/(i*i), i=1, 1000) /)
real :: result

result = sum(a);

Or in Fortran 77:

      s=0
      do i=1,1000
          s=s+1./i**2
      end do
      write (*,*) s
      end

FreeBASIC

' FB 1.05.0 Win64

Const pi As Double = 3.141592653589793

Function sumSeries (n As UInteger) As Double
  If n = 0 Then Return 0
  Dim sum As Double = 0
  For k As Integer = 1 To n
    sum += 1.0/(k * k)
  Next
  Return sum
End Function

Print "s(1000) = "; sumSeries(1000)
Print "zeta(2) = "; Pi * pi / 6
Print
Print "Press any key to quit"
Sleep

{{out}}

s(1000) =  1.643934566681562
zeta(2) =  1.644934066848226

Frink

Frink can calculate the series with exact rational numbers or floating-point values.

sum[map[{|k| 1/k^2}, 1 to 1000]]

{{out}}

83545938483149689478187854264854884386044454314086472930763839512603803291207881839588904977469387999844962675327115010933903589145654299730231109091124308462732153297321867661093162618281746011828755017021645889046777854795025297006943669294330752479399654716368801794529682603741344724733173765262964463970763934463926259796895140901128384286333311745462863716753134735154188954742414035836608258393970996630553795415075904205673610359458498106833291961256452756993199997231825920203667952667546787052535763624910912251107083702817265087341966845358732584971361645348091123849687614886682117125784781422103460192439394780707024963279033532646857677925648889105430050030795563141941157379481719403833258405980463950499887302926152552848089894630843538497552630691676216896740675701385847032173192623833881016332493844186817408141003602396236858699094240207812766449/50820720104325812617835292273000760481839790754374852703215456050992581046448162621598030244504097240825920773913981926305208272518886258627010933716354037062979680120674828102224650586465553482032614190502746121717248161892239954030493982549422690846180552358769564169076876408783086920322038142618269982747137757706040198826719424371333781947889528085329853597116893889786983109597085041878513917342099206896166585859839289193299599163669641323895022932959750057616390808553697984192067774252834860398458100840611325353202165675189472559524948330224159123505567527375848194800452556940453530457590024173749704941834382709198515664897344438584947842793131829050180589581507273988682409028088248800576590497216884808783192565859896957125449502802395453976401743504938336291933628859306247684023233969172475385327442707968328512729836445886537101453118476390400000000 (approx. 1.6439345666815598)

Change 1/k^2 to 1.0/k^2 to use floating-point math.

=={{header|F_Sharp|F#}}== The following function will do the task specified.

let rec f (x : float) =
    match x with
        | 0. -> x
        | x -> (1. / (x * x)) + f (x - 1.)

In the interactive F# console, using the above gives:

 f 1000. ;;
val it : float = 1.643934567

However this recursive function will run out of stack space eventually (try 100000). A [[:Category:Recursion|tail-recursive]] implementation will not consume stack space and can therefore handle much larger ranges. Here is such a version:

#light
let sum_series (max : float) =
    let rec f (a:float, x : float) =
        match x with
            | 0. -> a
            | x -> f ((1. / (x * x) + a), x - 1.)
    f (0., max)

[<EntryPoint>]
let main args =
    let (b, max) = System.Double.TryParse(args.[0])
    printfn "%A" (sum_series max)
    0

This block can be compiled using ''fsc --target exe filename.fs'' or used interactively without the main function.

GAP

# We will compute the sum exactly

# Computing an approximation of a rationnal (giving a string)
# Value is truncated toward zero
Approx := function(x, d)
	local neg, a, b, n, m, s;
	if x < 0 then
		x := -x;
		neg := true;
	else
		neg := false;
	fi;
	a := NumeratorRat(x);
	b := DenominatorRat(x);
	n := QuoInt(a, b);
	a := RemInt(a, b);
	m := 10^d;
	s := "";
	if neg then
		Append(s, "-");
	fi;
	Append(s, String(n));
	n := Size(s) + 1;
	Append(s, String(m + QuoInt(a*m, b)));
	s[n] := '.';
	return s;
end;

a := Sum([1 .. 1000], n -> 1/n^2);;
Approx(a, 10);
"1.6439345666"
# and pi^2/6 is 1.6449340668, truncated to ten digits

Genie

[indent=4]
/*
   Sum of series, in Genie
   valac sumOfSeries.gs
   ./sumOfSeries
*/

delegate sumFunc(n:int):double

def sum_series(start:int, end:int, f:sumFunc):double
    sum:double = 0.0
    for var i = start to end do sum += f(i)
    return sum


def oneOverSquare(n:int):double
    return (1 / (double)(n * n))

init
    Intl.setlocale()
    print "ζ(2) approximation: %16.15f", sum_series(1, 1000, oneOverSquare)
    print "π² / 6            : %16.15f", Math.PI * Math.PI / 6.0

{{out}}

prompt$ valac sumOfSeries.gs
prompt$ ./sumOfSeries
ζ(2) approximation: 1.643934566681561
π² / 6            : 1.644934066848226

GEORGE

0 (s)
1, 1000 rep (i)
   s 1 i dup × / + (s) ;
]
P

Output:-

 1.643934566681561

Go

package main

import ("fmt"; "math")

func main() {
    fmt.Println("known:   ", math.Pi*math.Pi/6)
    sum := 0.
    for i := 1e3; i > 0; i-- {
        sum += 1 / (i * i)
    }
    fmt.Println("computed:", sum)
}

Output:

known:    1.6449340668482264
computed: 1.6439345666815597

Groovy

Start with smallest terms first to minimize rounding error:

println ((1000..1).collect { x -> 1/(x*x) }.sum())

Output:

1.6439345654

Haskell

With a list comprehension:

sum [1 / x ^ 2 | x <- [1..1000]]

With higher-order functions:

sum $ map (\x -> 1 / x ^ 2) [1..1000]

In [http://haskell.org/haskellwiki/Pointfree point-free] style:

(sum . map (1/) . map (^2)) [1..1000]

or

(sum . map ((1 /) . (^ 2))) [1 .. 1000]

or, as a single fold:

seriesSum f = foldr ((+) . f) 0

inverseSquare = (1 /) . (^ 2)

main :: IO ()
main = print $ seriesSum inverseSquare [1 .. 1000]

{{Out}}

1.6439345666815615

HicEst

REAL :: a(1000)
        a = 1 / $^2
        WRITE(ClipBoard, Format='F17.15') SUM(a)

=={{header|Icon}} and {{header|Unicon}}==

```icon
procedure main()
   local i, sum
   sum := 0 & i := 0
   every sum +:= 1.0/((| i +:= 1 ) ^ 2) \1000
   write(sum)
end

or

procedure main()
    every (sum := 0) +:= 1.0/((1 to 1000)^2)
    write(sum)
end

Note: The terse version requires some explanation. Icon expressions all return values or references if they succeed. As a result, it is possible to have expressions like these below:

   x := y := 0   # := is right associative so, y is assigned 0, then x
   1 < x < 99    # comparison operators are left associative so, 1 < x returns x (if it is greater than 1), then x < 99 returns 99 if the comparison succeeds
   (sum := 0)    # returns a reference to sum which can in turn be used with augmented assignment +:=

IDL

print,total( 1/(1+findgen(1000))^2)

Io

Io 20110905
Io> sum := 0 ; Range 1 to(1000) foreach(k, sum = sum + 1/(k*k))
==> 1.6439345666815615
Io> 1 to(1000) map(k, 1/(k*k)) sum
==> 1.6439345666815615
Io>

The expression using map generates a list internally. Using foreach does not.

J

   NB. sum of reciprocals of squares of first thousand positive integers
   +/ % *: >: i. 1000
1.64393

   (*:o.1)%6       NB. pi squared over six, for comparison
1.64493

   1r6p2           NB.  As a constant (J has a rich constant notation)
1.64493

Java

public class Sum{
    public static double f(double x){
       return 1/(x*x);
    }

    public static void main(String[] args){
       double start = 1;
       double end = 1000;
       double sum = 0;

       for(double x = start;x <= end;x++) sum += f(x);

       System.out.println("Sum of f(x) from " + start + " to " + end +" is " + sum);
    }
}

JavaScript

ES5

function sum(a,b,fn) {
   var s = 0;
   for ( ; a <= b; a++) s += fn(a);
   return s;
}

 sum(1,1000, function(x) { return 1/(x*x) } )  // 1.64393456668156

or, in a functional idiom:

(function () {

  function sum(fn, lstRange) {
    return lstRange.reduce(
      function (lngSum, x) {
        return lngSum + fn(x);
      }, 0
    );
  }

  function range(m, n) {
    return Array.apply(null, Array(n - m + 1)).map(function (x, i) {
      return m + i;
    });
  }


  return sum(
    function (x) {
      return 1 / (x * x);
    },
    range(1, 1000)
  );

})();

{{Out}}

### ES6

{{Trans|Haskell}}

```JavaScript
(() => {
    'use strict';

    // SUM OF A SERIES -------------------------------------------------------

    // seriesSum :: Num a => (a -> a) -> [a] -> a
    const seriesSum = (f, xs) =>
        foldl((a, x) => a + f(x), 0, xs);


    // GENERIC ---------------------------------------------------------------

    // enumFromToInt :: Int -> Int -> [Int]
    const enumFromTo = (m, n) =>
        Array.from({
            length: Math.floor(n - m) + 1
        }, (_, i) => m + i);

    // foldl :: (b -> a -> b) -> b -> [a] -> b
    const foldl = (f, a, xs) => xs.reduce(f, a);

    // TEST ------------------------------------------------------------------

    return seriesSum(x => 1 / (x * x), enumFromTo(1, 1000));
})();

{{Out}}

## jq

The jq idiom for efficient computation of this kind of sum is to use "reduce", either directly or using a summation wrapper function.

Directly:

```jq
def s(n): reduce range(1; n+1) as $k (0; . + 1/($k * $k) );

s(1000)

{{Out}} 1.6439345666815615

Using a generic summation wrapper function allows problems specified in "sigma" notation to be solved using syntax that closely resembles that notation:

def summation(s): reduce s as $k (0; . + $k);

summation( range(1; 1001) | (1/(. * .) ) )

An important point is that nothing is lost in efficiency using the declarative and quite elegant approach using "summation".

Jsish

From Javascript ES5.

#!/usr/bin/jsish
/* Sum of a series */
function sum(a:number, b:number , fn:function):number {
   var s = 0;
   for ( ; a <= b; a++) s += fn(a);
   return s;
}

;sum(1, 1000, function(x) { return 1/(x*x); } );

/*
=!EXPECTSTART!=
sum(1, 1000, function(x) { return 1/(x*x); } ) ==> 1.643934566681561
=!EXPECTEND!=
*/

{{out}}

prompt$ jsish --U sumOfSeries.jsi
sum(1, 1000, function(x) { return 1/(x*x); } ) ==> 1.643934566681561

Julia

Using a higher-order function:

 sum(k -> 1/k^2, 1:1000)
1.643934566681559

julia> pi^2/6
1.6449340668482264

A simple loop is more optimized:

 function f(n)
    s = 0.0
    for k = 1:n
      s += 1/k^2
    end
    return s
end

julia> f(1000)
1.6439345666815615

K

  ssr: +/1%_sqr
  ssr 1+!1000
1.643935

Kotlin

// version 1.0.6

fun main(args: Array<String>) {
    val n = 1000
    val sum = (1..n).sumByDouble { 1.0 / (it * it) }
    println("Actual sum is $sum")
    println("zeta(2)    is ${Math.PI * Math.PI / 6.0}")
}

{{out}}

Actual sum is 1.6439345666815615
zeta(2)    is 1.6449340668482264

Lang5

1000 iota 1 + 1 swap / 2 ** '+ reduce .

Lasso

define sum_of_a_series(n::integer,k::integer) => {
	local(sum = 0)
	loop(-from=#k,-to=#n) => {
		#sum += 1.00/(math_pow(loop_count,2))
	}
	return #sum
}
sum_of_a_series(1000,1)

{{out}}

1.643935

LFE

With lists:foldl

(defun sum-series (nums)
  (lists:foldl
    #'+/2
    0
    (lists:map
      (lambda (x) (/ 1 x x))
      nums)))

With lists:sum

(defun sum-series (nums)
  (lists:sum
    (lists:map
      (lambda (x) (/ 1 x x))
      nums)))

Both have the same result:

> (sum-series (lists:seq 1 100000))
1.6449240668982423

Liberty BASIC

for i =1 to 1000
  sum =sum +1 /( i^2)
next i

print sum

end

Lingo

the floatprecision = 8
sum = 0
repeat with i = 1 to 1000
  sum = sum + 1/power(i, 2)
end repeat
put sum
-- 1.64393457

LiveCode

repeat with i = 1 to 1000
    add 1/(i^2) to summ
end repeat
put summ  //1.643935

Logo

to series :fn :a :b
  localmake "sigma 0
  for [i :a :b] [make "sigma :sigma + invoke :fn :i]
  output :sigma
end
to zeta.2 :x
  output 1 / (:x * :x)
end
print series "zeta.2 1 1000
make "pi (radarctan 0 1) * 2
print :pi * :pi / 6

Lua

sum = 0
for i = 1, 1000 do sum = sum + 1/i^2 end
print(sum)

Lucid

= 1000
   where
         num = 1 fby num + 1;
         ssum = ssum + 1/(num * num)
   end;

Maple

sum(1/k^2, k=1..1000);

{{Out|Output}}

-Psi(1, 1001)+(1/6)*Pi^2

Mathematica

This is the straightforward solution of the task:

Sum[1/x^2, {x, 1, 1000}]

However this returns a quotient of two huge integers (namely the ''exact'' sum); to get a floating point approximation, use N:

N[Sum[1/x^2, {x, 1, 1000}]]

or better:

NSum[1/x^2, {x, 1, 1000}]

Which gives a higher or equal accuracy/precision. Alternatively, get Mathematica to do the whole calculation in floating point by using a floating point value in the formula:

Sum[1./x^2, {x, 1, 1000}]

Other ways include (exact, approximate,exact,approximate):

Total[Table[1/x^2, {x, 1, 1000}]]
Total[Table[1./x^2, {x, 1, 1000}]]
Plus@@Table[1/x^2, {x, 1, 1000}]
Plus@@Table[1./x^2, {x, 1, 1000}]

MATLAB

   sum([1:1000].^(-2))

Maxima

(%i45) sum(1/x^2, x, 1, 1000);
       835459384831496894781878542648[806 digits]396236858699094240207812766449
(%o45) ------------------------------------------------------------------------
       508207201043258126178352922730[806 digits]886537101453118476390400000000

(%i46) sum(1/x^2, x, 1, 1000),numer;
(%o46) 1.643934566681561

MAXScript

total = 0
for i in 1 to 1000 do
(
    total += 1.0 / pow i 2
)
print total

min

{{works with|min|0.19.3}}

0 1 (
  ((dup * 1 swap /) (id)) cleave
  ((+) (succ)) spread
) 1000 times pop print

{{out}}

1.643934566681562

MiniScript

zeta = function(num)
    return 1 / num^2
end function

sum = function(start, finish, formula)
    total = 0
    for i in range(start, finish)
        total = total + formula(i)
    end for
    return total
end function

print sum(1, 1000, @zeta)

{{out}}

1.643935

=={{header|MK-61/52}}== 0 П0 П1 ИП1 1 + П1 x^2 1/x ИП0

• П0 ИП1 1 0 0 0 - x>=0 03 ИП0 С/П

## ML


=
## Standard ML
=

```Standard ML

(* 1.64393456668 *)
List.foldl op+ 0.0 (List.tabulate(1000, fn x => 1.0 / Math.pow(real(x + 1),2.0)))

=

mLite

=

println ` fold (+, 0) ` map (fn x = 1 / x ^ 2) ` iota (1,1000);

Output:

1.6439345666815549

MMIX

x	IS	$1	% flt calculations
y	IS	$2	%   id
z	IS	$3	% z = sum series
t	IS	$4	% temp var

	LOC	Data_Segment
	GREG	@
BUF	OCTA	0,0,0		% print buffer

	LOC	#1000
	GREG	@

// print floating point number in scientific format: 0.xxx...ey..
// most of this routine is adopted from:
// http://www.pspu.ru/personal/eremin/emmi/rom_subs/printreal.html
// float number in z
	GREG	@
NaN	BYTE	"NaN..",0
NewLn	BYTE	#a,0
1H	LDA	x,NaN
	TRAP	0,Fputs,StdOut
	GO	$127,$127,0

prtFlt	FUN	x,z,z		% test if z == NaN
	BNZ	x,1B
	CMP	$73,z,0		% if necessary remember it is neg
	BNN	$73,4F
Sign	BYTE	'-'
	LDA	$255,Sign
	TRAP	0,Fputs,StdOut
	ANDNH	z,#8000		% make number pos
// normalizing float number
4H	SETH	$74,#4024	% initialize mulfactor = 10.0
	SETH	$73,#0023
	INCMH	$73,#86f2
	INCML	$73,#6fc1	%
	FLOT	$73,$73		% $73 = float 10^16
	SET	$75,16		% set # decimals to 16
8H	FCMP	$72,z,$73	% while z >= 10^16 do
	BN	$72,9F		%
	FDIV	z,z,$74		%  z = z / 10.0
	ADD	$75,$75,1	%  incr exponent
	JMP	8B		% wend
9H	FDIV	$73,$73,$74	% 10^16 / 10.0
5H	FCMP	$72,z,$73	% while z < 10^15 do
	BNN	$72,6F
	FMUL	z,z,$74		%  z = z * 10.0
	SUB	$75,$75,1	%  exp = exp - 1
	JMP	5B
NulPnt	BYTE	'0','.',#00
6H	LDA	$255,NulPnt	% print '0.' to StdOut
	TRAP	0,Fputs,StdOut
	FIX	z,0,z		% convert float z to integer
// print mantissa
0H	GREG	#3030303030303030
	STO	0B,BUF
	STO	0B,BUF+8	% store print mask in buffer
	LDA	$255,BUF+16	% points after LSD
				% repeat
2H	SUB	$255,$255,1	%   move pointer down
	DIV	z,z,10		%   (q,r) = divmod z 10
	GET	t,rR		%   get remainder
	INCL	t,'0'		%   convert to ascii digit
	STBU	t,$255,0	%   store digit in buffer
	BNZ	z,2B		% until q == 0
	TRAP	0,Fputs,StdOut	% print mantissa
Exp	BYTE	'e',#00
	LDA	$255,Exp	% print 'exponent' indicator
	TRAP	0,Fputs,StdOut
// print exponent
0H	GREG	#3030300000000000
	STO	0B,BUF
	LDA	$255,BUF+2	% store print mask in buffer
	CMP	$73,$75,0	% if exp neg then place - in buffer
	BNN	$73,2F
ExpSign	BYTE	'-'
	LDA	$255,ExpSign
	TRAP	0,Fputs,StdOut
	NEG	$75,$75		% make exp positive
2H	LDA	$255,BUF+3	% points after LSD
				% repeat
3H	SUB	$255,$255,1	%   move pointer down
	DIV	$75,$75,10	%   (q,r) = divmod exp 10
	GET	t,rR
	INCL	t,'0'
	STBU	t,$255,0	%   store exp. digit in buffer
	BNZ	$75,3B		% until q == 0
	TRAP	0,Fputs,StdOut	% print exponent
	LDA	$255,NewLn
	TRAP	0,Fputs,StdOut	% do a NL
	GO	$127,$127,0	% return

i  IS $5 ;iu IS $6
Main	SET	iu,1000
	SETH	y,#3ff0     y = 1.0
	SETH	z,#0000     z = 0.0
	SET	i,1          for (i=1;i<=1000; i++ ) {
1H	FLOT	x,i           x = int i
	FMUL	x,x,x         x = x^2
	FDIV	x,y,x         x = 1 / x
	FADD	z,z,x         s = s + x
	ADD	i,i,1
	CMP	t,i,iu
	PBNP	t,1B         } z = sum
	GO	$127,prtFlt  print sum --> StdOut
	TRAP	0,Halt,0

Output:

~/MIX/MMIX/Rosetta> mmix sumseries
0.1643934566681562e1

=={{header|Modula-3}}== Modula-3 uses D0 after a floating point number as a literal for LONGREAL.

MODULE Sum EXPORTS Main;

IMPORT IO, Fmt, Math;

VAR sum: LONGREAL := 0.0D0;

PROCEDURE F(x: LONGREAL): LONGREAL =
  BEGIN
    RETURN 1.0D0 / Math.pow(x, 2.0D0);
  END F;

BEGIN
  FOR i := 1 TO 1000 DO
    sum := sum + F(FLOAT(i, LONGREAL));
  END;
  IO.Put("Sum of F(x) from 1 to 1000 is ");
  IO.Put(Fmt.LongReal(sum));
  IO.Put("\n");
END Sum.

Output:

Sum of F(x) from 1 to 1000 is 1.6439345666815612

MUMPS

SOAS(N)
 NEW SUM,I SET SUM=0
 FOR I=1:1:N DO
 .SET SUM=SUM+(1/((I*I)))
 QUIT SUM

This is an extrinsic function so the usage is:

USER>SET X=$$SOAS^ROSETTA(1000) WRITE X
1.643934566681559806

Nial

|sum (1 / power (count 1000) 2)
=1.64393

NewLISP

(let (s 0)
  (for (i 1 1000)
    (inc s (div 1 (* i i))))
  (println s))

Nim

import math

var ls: seq[float] = @[]
for x in 1..1000:
  ls.add(1.0 / float(x * x))
echo sum(ls)

Objeck

bundle Default {
  class SumSeries {
    function : Main(args : String[]) ~ Nil {
      DoSumSeries();
    }

    function : native : DoSumSeries() ~ Nil {
      start := 1;
      end := 1000;

      sum := 0.0;

      for(x : Float := start; x <= end; x += 1;) {
        sum += f(x);
      };

      IO.Console->GetInstance()->Print("Sum of f(x) from ")->Print(start)->Print(" to ")->Print(end)->Print(" is ")->PrintLine(sum);
    }

    function : native : f(x : Float) ~ Float {
      return 1.0 / (x * x);
    }
  }
}

OCaml

let sum a b fn =
  let result = ref 0. in
  for i = a to b do
    result := !result +. fn i
  done;
  !result

sum 1 1000 (fun x -> 1. /. (float x ** 2.))

• : float = 1.64393456668156124

or in a functional programming style:

let sum a b fn =
  let rec aux i r =
    if i > b then r
    else aux (succ i) (r +. fn i)
  in
  aux a 0.
;;

Simple recursive solution:

let rec sum n = if n < 1 then 0.0 else sum (n-1) +. 1.0 /. float (n*n)
in sum 1000

Octave

Given a vector, the sum of all its elements is simply sum(vector); a range can be ''generated'' through the range notation: sum(1:1000) computes the sum of all numbers from 1 to 1000. To compute the requested series, we can simply write:

sum(1 ./ [1:1000] .^ 2)

Oforth

: sumSerie(s, n)   0 n seq apply(#[ s perform + ]) ;

Usage :

 #[ sq inv ] 1000 sumSerie println

{{out}}

1.64393456668156

OpenEdge/Progress

Conventionally like elsewhere:

<lang Progress (Openedge ABL)>def var dcResult as decimal no-undo. def var n as int no-undo.

do n = 1 to 1000 : dcResult = dcResult + 1 / (n * n) . end.

display dcResult .

or like this:

<lang Progress (Openedge ABL)>def var n as int no-undo.

repeat n = 1 to 1000 :
  accumulate 1 / (n * n) (total).
end.

display ( accum total 1 / (n * n) )  .

Oz

With higher-order functions:

declare
  fun {SumSeries S N}
     {FoldL {Map {List.number 1 N 1} S}
      Number.'+' 0.}
  end

  fun {S X}
     1. / {Int.toFloat X*X}
  end
in
  {Show {SumSeries S 1000}}

Iterative:

  fun {SumSeries S N}
     R = {NewCell 0.}
  in
     for I in 1..N do
        R := @R + {S I}
     end
     @R
  end

PARI/GP

Exact rational solution:

sum(n=1,1000,1/n^2)

Real number solution (accurate to $3\backslash cdot10^\{-36\}$ at standard precision):

sum(n=1,1000,1./n^2)

Approximate solution (accurate to $9\backslash cdot10^\{-11\}$ at standard precision):

zeta(2)-intnum(x=1000.5,[1],1/x^2)

or

zeta(2)-1/1000.5

Panda

sum{{1.0.divide(1..1000.sqr)}}

Output:

1.6439345666815615

Pascal

Program SumSeries;
type
  tOutput = double;//extended;
  tmyFunc = function(number: LongInt): tOutput;

function f(number: LongInt): tOutput;
begin
  f := 1/sqr(tOutput(number));
end;

function Sum(from,upto: LongInt;func:tmyFunc):tOutput;
var
  res: tOutput;
begin
  res := 0.0;
//  for from:= from to upto do res := res + f(from);
  for upTo := upto downto from do res := res + f(upTo);
  Sum := res;
end;

BEGIN
  writeln('The sum of 1/x^2 from 1 to 1000 is: ', Sum(1,1000,@f));
  writeln('Whereas pi^2/6 is:                  ', pi*pi/6:10:8);
end.

Output

different version of type and calculation
extended low to high 1.64393456668155980263E+0000
extended high to low 1.64393456668155980307E+0000
  double low to high 1.6439345666815612E+000
  double high to low 1.6439345666815597E+000
Out:
The sum of 1/x^2 from 1 to 1000 is:  1.6439345666815612E+000
Whereas pi^2/6 is:                  1.64493407

Perl

my $sum = 0;
$sum += 1 / $_ ** 2 foreach 1..1000;
print "$sum\n";

or

use List::Util qw(reduce);
$sum = reduce { $a + 1 / $b ** 2 } 0, 1..1000;
print "$sum\n";

An other way of doing it is to define the series as a closure:

my $S = do { my ($sum, $k); sub { $sum += 1/++$k**2 } };
my @S = map &$S, 1 .. 1000;
print $S[-1];

Perl 6

{{Works with|rakudo|2016.04}}

In general, the $nth partial sum of a series whose terms are given by a unary function &f is

[+] map &f, 1 .. $n

So what's needed in this case is

say [+] map { 1 / $^n**2 }, 1 .. 1000;

Or, using the "hyper" metaoperator to vectorize, we can use a more "point free" style while keeping traditional precedence:

say [+] 1 «/« (1..1000) »**» 2;

Or we can use the X "cross" metaoperator, which is convenient even if one side or the other is a scalar. In this case, we demonstrate a scalar on either side:

say [+] 1 X/ (1..1000 X** 2);

Note that cross ops are parsed as list infix precedence rather than using the precedence of the base op as hypers do. Hence the difference in parenthesization.

With list comprehensions, you can write:

say [+] (1 / $_**2 for 1..1000);

That's fine for a single result, but if you're going to be evaluating the sequence multiple times, you don't want to be recalculating the sum each time, so it's more efficient to define the sequence as a constant to let the run-time automatically cache those values already calculated. In a lazy language like Perl 6, it's generally considered a stronger abstraction to write the correct infinite sequence, and then take the part of it you're interested in. Here we define an infinite sequence of partial sums (by adding a backslash into the reduction to make it look "triangular"), then take the 1000th term of that:

constant @x = [\+] 0, { 1 / ++(state $n) ** 2 } ... *;
say @x[1000];  # prints 1.64393456668156

Note that infinite constant sequences can be lazily generated in Perl 6, or this wouldn't work so well...

A cleaner style is to combine these approaches with a more FP look:

constant ζish = [\+] map -> \𝑖 { 1 / 𝑖**2 }, 1..*;
say ζish[1000];

Perhaps the cleanest way is to just define the zeta function and evaluate it for s=2, possibly using memoization:

use experimental :cached;
sub ζ($s) is cached { [\+] 1..* X** -$s }
say ζ(2)[1000];

Notice how the thus-defined zeta function returns a lazy list of approximated values, which is arguably the closest we can get from the mathematical definition.

Phix

function sumto(atom n)
atom res = 0
    for i=1 to n do
        res += 1/(i*i)
    end for
    return res
end function
?sumto(1000)

{{out}}

1.643934567

PHP

<?php

/**
 * @author Elad Yosifon
 */

/**
 * @param int $n
 * @param int $k
 * @return float|int
 */
function sum_of_a_series($n,$k)
{
	$sum_of_a_series = 0;
	for($i=$k;$i<=$n;$i++)
	{
		$sum_of_a_series += (1/($i*$i));
	}
	return $sum_of_a_series;
}

echo sum_of_a_series(1000,1);

{{out}}

1.6439345666816

PicoLisp

(scl 9)  # Calculate with 9 digits precision

(let S 0
   (for I 1000
      (inc 'S (*/ 1.0 (* I I))) )
   (prinl (round S 6)) )  # Round result to 6 digits

Output:

1.643935

Pike

array(int) x = enumerate(1000,1,1);
`+(@(1.0/pow(x[*],2)[*]));
Result: 1.64393

PL/I

/* sum the first 1000 terms of the series 1/n**2. */
s = 0;

do i = 1000 to 1 by -1;
   s = s + 1/float(i**2);
end;

put skip list (s);

{{out}}

1.64393456668155980E+0000

Pop11

lvars s = 0, j;
for j from 1 to 1000 do
    s + 1.0/(j*j) -> s;
endfor;

s =>

PostScript

Output:
<lang>
1.64393485

% using map
[1 1000] 1 range {dup * 1 exch div} map 0 {+} fold

% just using fold
[1 1000] 1 range 0 {dup * 1 exch div +}  fold

sum = 0.0
1 to 1000 (i): sum = sum + 1.0 / (i * i).
sum print

$x = 1..1000 `
       | ForEach-Object { 1 / ($_ * $_) } `
       | Measure-Object -Sum
Write-Host Sum = $x.Sum

sum(S) :-
        findall(L, (between(1,1000,N),L is 1/N^2), Ls),
        sumlist(Ls, S).

?- sum(S).
S = 1.643934566681562.

Define i, sum.d

For i=1 To 1000
  sum+1.0/(i*i)
Next i

Debug sum

print ( sum(1.0 / (x * x) for x in range(1, 1001)) )

'''The sum of a series'''

from functools import reduce


# seriesSumA :: (a -> b) -> [a] -> b
def seriesSumA(f):
    '''The sum of the map of f over xs.'''
    return lambda xs: sum(map(f, xs))


# seriesSumB :: (a -> b) -> [a] -> b
def seriesSumB(f):
    '''Folding acc + f(x) over xs where acc begins at 0.'''
    return lambda xs: reduce(
        lambda a, x: a + f(x), xs, 0
    )


# TEST ----------------------------------------------------
# main:: IO ()
def main():
    '''Summing 1/x^2 over x = 1..1000'''

    def f(x):
        return 1 / (x * x)

    print(
        fTable(
            __doc__ + ':\n' + '(1/x^2 over x = 1..1000)'
        )(lambda f: '\tby ' + f.__name__)(str)(
            lambda g: g(f)(enumFromTo(1)(1000))
        )([seriesSumA, seriesSumB])
    )


# GENERIC -------------------------------------------------

# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
    '''Right to left function composition.'''
    return lambda f: lambda x: g(f(x))


# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
    '''Integer enumeration from m to n.'''
    return lambda n: list(range(m, 1 + n))


# fTable :: String -> (a -> String) ->
#                     (b -> String) ->
#        (a -> b) -> [a] -> String
def fTable(s):
    '''Heading -> x display function -> fx display function ->
          f -> value list -> tabular string.'''
    def go(xShow, fxShow, f, xs):
        w = max(map(compose(len)(xShow), xs))
        return s + '\n' + '\n'.join([
            xShow(x).rjust(w, ' ') + (
                ' -> '
            ) + fxShow(f(x)) for x in xs
        ])
    return lambda xShow: lambda fxShow: (
        lambda f: lambda xs: go(
            xShow, fxShow, f, xs
        )
    )


# MAIN ---
if __name__ == '__main__':
    main()

The sum of a series:
(1/x^2 over x = 1..1000)
    by seriesSumA -> 1.6439345666815615
    by seriesSumB -> 1.6439345666815615

sn:{sum xexp[;-2] 1+til x}
sn 1000

1.643935

print( sum( 1/seq(1000)^2 ) )

#lang typed/racket

(: S : Natural -> Real)
(define (S n)
  (for/sum: : Real ([k : Natural (in-range 1 (+ n 1))])
    (/ 1.0 (* k k))))

0 1 1000 1 range each 1.0 swap dup * / +
"%g\n" print

1.64393

Red []
s: 0
repeat n 1000 [  s:   1.0 / n ** 2  + s  ]
print s

/*REXX program sums the first    N    terms of     1/(k**2),          k=1 ──►  N.       */
parse arg N D .                                  /*obtain optional arguments from the CL*/
if N=='' | N==","  then N=1000                   /*Not specified?  Then use the default.*/
if D=='' | D==","  then D=  60                   /* "      "         "   "   "     "    */
numeric digits D                                 /*use D digits (9 is the REXX default).*/
$=0                                              /*initialize the sum to zero.          */
          do k=1  for N                          /* [↓]  compute for   N   terms.       */
          $=$  +  1/k**2                         /*add a squared reciprocal to the sum. */
          end   /*k*/

say 'The sum of'     N     "terms is:"    $      /*stick a fork in it,  we're all done. */

The sum of 1000 terms is: 1.64393456668155980313905802382221558965210344649368531671713

/*REXX program sums the first    N    terms o f    1/(k**2),          k=1 ──►  N.       */
parse arg N D .                                  /*obtain optional arguments from the CL*/
if N=='' | N==","  then N=1000                   /*Not specified?  Then use the default.*/
if D=='' | D==","  then D=  60                   /* "      "         "   "   "     "    */
numeric digits D                                 /*use D digits (9 is the REXX default).*/
w=length(N)                                      /*W   is used for aligning the output. */
$=0                                              /*initialize the sum to zero.          */
      do k=1  for N                              /* [↓]  compute for   N   terms.       */
      $=$  +  1/k**2                             /*add a squared reciprocal to the sum. */
      parse var k s 2 m '' -1 e                  /*obtain the start and end decimal digs*/
      if e\==0  then iterate                     /*does K  end  with the dec digit  0 ? */
      if s\==1  then iterate                     /*  "  " start   "   "   "    "    1 ? */
      if m\=0   then iterate                     /*  "  " middle  contain any non-zero ?*/
      if k==N   then iterate                     /*  "  " equal N, then skip running sum*/
      say  'The sum of'   right(k,w)     "terms is:"  $         /*display a running sum.*/
      end   /*k*/
say                                                             /*a blank line for sep. */
say        'The sum of'   right(k-1,w)   "terms is:"  $         /*display the final sum.*/
                                                 /*stick a fork in it,  we're all done. */

The sum of         10 terms is: 1.54976773116654069035021415973796926177878558830939783320736
The sum of        100 terms is: 1.63498390018489286507716949818032376668332170003126381385307
The sum of       1000 terms is: 1.64393456668155980313905802382221558965210344649368531671713
The sum of      10000 terms is: 1.64483407184805976980608183331031090353799751949684175308996
The sum of     100000 terms is: 1.64492406689822626980574850331269185564752132981156034248806
The sum of    1000000 terms is: 1.64493306684872643630574849997939185588561654406394129491321
The sum of   10000000 terms is: 1.64493396684823143647224849997935852288561656787346272343397
The sum of  100000000 terms is: 1.64493405684822648647241499997935852255228656787346510441026
The sum of 1000000000 terms is: 1.64493406584822643697241516647935852255228323457346510444171

/*REXX program sums the first    N    terms of     1/(k**2),          k=1 ──►  N.       */
parse arg N D .                                  /*obtain optional arguments from the CL*/
if N=='' | N==","  then N=1000                   /*Not specified?  Then use the default.*/
if D=='' | D==","  then D=  60                   /* "      "         "   "   "     "    */
numeric digits D                                 /*use D digits (9 is the REXX default).*/
w=length(N)                                      /*W   is used for aligning the output. */
$=0                                              /*initialize the sum to zero.          */
old=1                                            /*the new sum to compared to the old.  */
p=0                                              /*significant decimal precision so far.*/
     do k=1  for N                               /* [↓]  compute for   N   terms.       */
     $=$  +  1/k**2                              /*add a squared reciprocal to the sum. */
     c=compare($,old)                            /*see how we're doing with precision.  */
     if c>p  then do                             /*Got another significant decimal dig? */
                  say 'The significant sum of'  right(k,w)      "terms is:"      left($,c)
                  p=c                            /*use the new significant precision.   */
                  end                            /* [↑]  display significant part of sum*/
     old=$                                       /*use "old" sum for the next compare.  */
     end   /*k*/
say                                              /*display blank line for the separator.*/
say 'The sum of'   right(N,w)    "terms is:"     /*display the  sum's  preamble line.   */
say $                                            /*stick a fork in it,  we're all done. */

The significant sum of          3 terms is: 1.3
The significant sum of          5 terms is: 1.46
The significant sum of         14 terms is: 1.575
The significant sum of         34 terms is: 1.6159
The significant sum of        110 terms is: 1.63588
The significant sum of        328 terms is: 1.641889
The significant sum of       1024 terms is: 1.6439579
The significant sum of       3207 terms is: 1.64462229
The significant sum of      10043 terms is: 1.644834499
The significant sum of      31782 terms is: 1.6449026029
The significant sum of     100314 terms is: 1.64492409819
The significant sum of     316728 terms is: 1.644930909569
The significant sum of    1000853 terms is: 1.6449330677009
The significant sum of    3163463 terms is: 1.64493375073899
The significant sum of   10001199 terms is: 1.644933966860219
The significant sum of   31627592 terms is: 1.6449340352302649
The significant sum of  100009299 terms is: 1.64493405684915629
The significant sum of  316233759 terms is: 1.644934063686008709

The sum of 1000000000 terms is:
1.644934065848226436972415166479358522552283234573465104402224896012864613260343731009819376810240620

sum = 0
for i =1 to 1000
    sum = sum + 1 /(pow(i,2))
next
decimals(8)
see sum

>> sum( (1 ./ [1:1000]) .^ 2 ) - const.pi^2/6
-0.000999500167

puts (1..1000).inject{ |sum, x| sum + 1.0 / x ** 2 }

1.64393456668156

for i =1 to 1000
  sum = sum + 1 /( i^2)
next i
print sum

const LOWER: i32 = 1;
const UPPER: i32 = 1000;

// Because the rule for our series is simply adding one, the number of terms are the number of
// digits between LOWER and UPPER
const NUMBER_OF_TERMS: i32 = (UPPER + 1) - LOWER;
fn main() {
    // Formulaic method
    println!("{}", (NUMBER_OF_TERMS * (LOWER + UPPER)) / 2);
    // Naive method
    println!("{}", (LOWER..UPPER + 1).fold(0, |sum, x| sum + x));
}

data _null_;
s=0;
do n=1 to 1000;
   s+1/n**2;        /* s+x is synonym of s=s+x */
end;
e=s-constant('pi')**2/6;
put s e;
run;

 1 to 1000 map (x => 1.0 / (x * x)) sum
res30: Double = 1.6439345666815615

(define (sum a b fn)
  (do ((i a (+ i 1))
       (result 0 (+ result (fn i))))
      ((> i b) result)))

(sum 1 1000 (lambda (x) (/ 1 (* x x)))) ; fraction
(exact->inexact (sum 1 1000 (lambda (x) (/ 1 (* x x))))) ; decimal

(define (invsq f to)
  (let loop ((f f) (s 0))
    (if (> f to)
      s
      (loop (+ 1 f) (+ s (/ 1 f f))))))

;; whether you get a rational or a float depends on implementation
(invsq 1 1000) ; 835459384831...766449/50820...90400000000
(exact->inexact (invsq 1 1000)) ; 1.64393456668156

$ include "seed7_05.s7i";
  include "float.s7i";

const func float: invsqr (in float: n) is
  return 1.0 / n**2;

const proc: main is func
  local
    var integer: i is 0;
    var float: sum is 0.0;
  begin
    for i range 1 to 1000 do
      sum +:= invsqr(flt(i));
    end for;
    writeln(sum digits 6 lpad 8);
  end func;

say sum(1..1000, {|n| 1 / n**2 })

say (1..1000 -> reduce { |a,b| a + (1 / b**2) })

1.64393456668155980313905802382221558965210344649369

((1 to: 1000) reduce: [|:x :y | x + (y squared reciprocal as: Float)]).

( (1 to: 1000) fold: [:sum :aNumber |
  sum + (aNumber squared reciprocal) ] ) asFloat displayNl.

create table t1 (n real);
-- this is postgresql specific, fill the table
insert into t1 (select generate_series(1,1000)::real);
with tt as (
  select 1/(n*n) as recip from t1
) select sum(recip) from tt;

       sum
------------------
 1.64393456668156
(1 Zeile)

function series(n) {
	return(sum((n..1):^-2))
}

series(1000)-pi()^2/6
  -.0009995002

func sumSeries(var n: Int) -> Double {
    var ret: Double = 0

    for i in 1...n {
        ret += (1 / pow(Double(i), 2))
    }

    return ret
}

output: 1.64393456668156

    for i in 1...self {
       ret += (1 / pow(Double(i), 2))
    }

    return ret
}

## Tcl

{{works with|Tcl|8.5}}

```tcl
package require Tcl 8.5

proc partial_sum {func - start - stop} {
    for {set x $start; set sum 0} {$x <= $stop} {incr x} {
        set sum [expr {$sum + [apply $func $x]}]
    }
    return $sum
}

set S {x {expr {1.0 / $x**2}}}

partial_sum $S from 1 to 1000 ;# => 1.6439345666815615

package require Tcl 8.5
package require struct::list

proc sum_of {lambda nums} {
    struct::list fold [struct::list map $nums [list apply $lambda]] 0 ::tcl::mathop::+
}

sum_of $S [range 1 1001] ;# ==> 1.6439345666815615

# a range command akin to Python's
proc range args {
    foreach {start stop step} [switch -exact -- [llength $args] {
        1 {concat 0 $args 1}
        2 {concat   $args 1}
        3 {concat   $args  }
        default {error {wrong # of args: should be "range ?start? stop ?step?"}}
    }] break
    if {$step == 0} {error "cannot create a range when step == 0"}
    set range [list]
    while {$step > 0 ? $start < $stop : $stop < $start} {
        lappend range $start
        incr start $step
    }
    return $range
}

∑(1/X²,X,1,1000)

1.643934567

∑(1/x^2,x,1,1000)

txr -p '[reduce-left + (let ((i 0)) (gen (< i 1000) (/ 1.0 (* (inc i) i)))) 0]'
1.64393456668156

txr -p '[reduce-left + [(let ((i 0)) (gen t (/ 1.0 (* (inc i) i)))) 0..999] 0]'
1.64393456668156

txr -p '[[chain range (op mapcar (op / 1.0 (* @1 @1))) (op reduce-left + @1 0)] 1 1000]'
1.64393456668156

txr -p '[reduce-left + (mapcar (op / 1.0 (* @1 @1)) (range 1 1000)) 0]'
1.64393456668156

txr -p '[reduce-left (op + @1 (/ 1.0 (* @2 @2))) (range 1 1000) 0]'
1.64393456668156

term() {
   b=$1;res=$2
   echo "scale=5;1/($res*$res)+$b" | bc
}

sum() {
  (read B; res=$1;
  test -n "$B" && (term $B $res) || (term 0 $res))
}

fold() {
  func=$1
  (while read a ; do
      fold $func | $func $a
  done)
}

(echo 3; echo 1; echo 4) | fold sum

#import flo
#import nat

#cast %e

total = plus:-0 div/*1. sqr* float*t iota 1001

1.643935e+00

public static void main(){
	int i, start = 1, end = 1000;
	double sum = 0.0;

	for(i = start; i<= end; i++)
		sum += (1 / (double)(i * i));

	stdout.printf("%s\n", sum.to_string());
}

1.6439345666815615

Private Function sumto(n As Integer) As Double
    Dim res As Double
    For i = 1 To n
        res = res + 1 / i ^ 2
    Next i
    sumto = res
End Function
Public Sub main()
    Debug.Print sumto(1000)
End Sub

 1,64393456668156

' Sum of a series
    for i=1 to 1000
        s=s+1/i^2
    next
    wscript.echo s

1.64393456668156

' Sum of a series
    Sub SumOfaSeries()
        Dim s As Double
        s = 0
        For i = 1 To 1000
            s = s + 1 / i ^ 2
        Next 'i
        Console.WriteLine(s)
    End Sub

1.64393456668156

 import 'stream';

s.range 1 1001
-> s.map (@ inner k => / 1 (* k k))
-> s.reduce 0 +
-- io.writeln io.stdout
;

1.643933567

@sum !*#~V1Sn @to 1000 ; returns 1.6439345666815615

@to 1000 ; generates a list of 1 to 1000 (inclusive)
#~V1Sn ; number expression which stands for: square push(1) swap divide
!* ; maps the number expression over the list
@sum ; sums the list

code CrLf=9;  code real RlOut=48;
int  X;  real S;
[S:= 0.0;
for X:= 1 to 1000 do S:= S + 1.0/float(X*X);
RlOut(0, S);  CrLf(0);
]

    1.64393

(1./indgen(1:1000)^2)(sum)

[1.0..1000].reduce(fcn(p,n){ p + 1.0/(n*n) },0.0)  //-->1.64394