/*
 * 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.
 */

float simplesum (long n);

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

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