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

#include <stdio.h>
#include <math.h>

//#define N 1000000L
#define N 1000L
#define NREP 10

double simplesum (long n);

int main (void)
{
    double s;
    int k;
    long n = N;

    for (k = 0; k <= NREP; k++)
    {
        s = simplesum (n);
        printf ("%20ld   %20.14f   %20.14f\n", n, s, fabs (0.5 - s));
        n *= 2;
    }

    return (0);
}