/*
 * 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
 *
 * Use Cahan summation algorithm.
 */

#include <stdio.h>

#define N 1000000
#define NREP 12

double gregory(long n) 
{
    double s = 0., snew, del, cor, cors = 0.;
    long k;

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

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

    for (k = 0; k <= NREP; k++) 
    {
        s = gregory(n);
        printf("%20ld   %20.14f   %20.14f\n", n, s, 1./2. - s);
        n *= 2;
    }


    return(0);
}