Compute the   nth   term of a series,   i.e. the sum of the   n   first terms of the corresponding 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


=

∑

k
=
1


n




1

k

2






{\displaystyle S_{n}=\sum _{k=1}^{n}{\frac {1}{k^{2}}}}




 and compute   




S

1000




{\displaystyle S_{1000}}




This approximates the   zeta function   for   S=2,   whose exact value

 



ζ
(
2
)
=



π

2


6




{\displaystyle \zeta (2)={\pi ^{2} \over 6}}


is the solution of the Basel problem.




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);
    }
}