/*
 * Calculate the sum: sum_{k=1}^n 1/(4k^2 - 1) =
 *  sum{k=1}^\n 1/2*(1/(2*k - 1) - 1/(2*k + 1)) = 1/2 (1 - 1/(2*n + 1))
 *
 * Naive implementation.
 */

double simplesum (long n) 
{
    double s = 0.;
    long k;

    for (k = 1; k <= n; k++) 
    {
        s += 1./((double) (4*k*k - 1));
    }
    return(s);
}