Example of accelerating a series

The following code calculates an estimate of @math{\zeta(2) = \pi^2 / 6} using the series,

After N terms the error in the sum is @math{O(1/N)}, making direct summation of the series converge slowly.

#include <stdio.h>
#include <gsl/gsl_math.h>
#include <gsl/gsl_sum.h>

#define N 20

int
main (void)
{
  double t[N];
  double sum_accel, err;
  double sum = 0;
  int n;
  
  gsl_sum_levin_u_workspace * w 
    = gsl_sum_levin_u_alloc (N);

  const double zeta_2 = M_PI * M_PI / 6.0;
  
  /* terms for zeta(2) = \sum_{n=0}^{\infty} 1/n^2 */

  for (n = 0; n < N; n++)
    {
      double np1 = n + 1.0;
      t[n] = 1.0 / (np1 * np1);
      sum += t[n];
    }
  
  gsl_sum_levin_u_accel (t, N, w, &sum_accel, &err);

  printf("term-by-term sum = % .16f using %d terms\n", 
         sum, N);

  printf("term-by-term sum = % .16f using %d terms\n", 
         w->sum_plain, w->terms_used);

  printf("exact value      = % .16f\n", zeta_2);
  printf("accelerated sum  = % .16f using %d terms\n", 
         sum_accel, w->terms_used);

  printf("estimated error  = % .16f\n", err);
  printf("actual error     = % .16f\n", 
         sum_accel - zeta_2);

  gsl_sum_levin_u_free (w);
  return 0;
}

The output below shows that the Levin @math{u}-transform is able to obtain an estimate of the sum to 1 part in @math{10^10} using the first eleven terms of the series. The error estimate returned by the function is also accurate, giving the correct number of significant digits.

bash$ ./a.out 
term-by-term sum =  1.5961632439130233 using 20 terms
term-by-term sum =  1.5759958390005426 using 13 terms
exact value      =  1.6449340668482264
accelerated sum  =  1.6449340668166479 using 13 terms
estimated error  =  0.0000000000508580
actual error     = -0.0000000000315785

Note that a direct summation of this series would require @math{10^10} terms to achieve the same precision as the accelerated sum does in 13 terms.