Sum of a series: Difference between revisions

For this task, use S(x) = 1/x^2, from 1 to 1000. (This approximates the Riemann zeta function. The Basel problem solved this: zeta(2) = πp²/6.)

unsigned int start = 1;

unsigned int end = 1000;

double sum = 0;

for( unsigned int x = start;

x <= end;

sum += f(x);

std::cout << "Sum of f(x) from " << start << " to " << end << " is " << sum << std::endl;

return 0;

return ( 1 / ( x * x ) );