As an example, the sample output below shows 4 acceptable lines out of 3001, which might appear in the output file. The values ofx should start at 0.00 and increase by 0.l until the line with x=300.00 is output. Sam ple Output(o represents a space) ■■0.00■■■1.644934066848 which implies y(1)=1.0. One can then ■■0.10■■■1534607244904 produce a series for yx)-yl) which converges faster than the original series. ■口100■■■10000000000 his series not only converges much faster. it also reduces roundoff loss. ■2.00■■■0.7500000000 his process of finding a faster converging series may be repeated 300.00■■■0.020942212934 again on the second series to produce a third sequence, which converges even To improve the convergence more rapidly than the second of the summation process The following equation is helpful in note that determining how may items are required in summing the series above k(k+1)kk+1 ∑k d r for k> I and r≥As an example, the sample output below shows 4 acceptable lines out of 3001, which might appear in the output file. The values of x should start at 0.00 and increase by 0.1 until the line with x = 300.00 is output. Sample Output ( represents a space) 0.001.644934066848 0.101.534607244904 ... 1.001.000000000000 ... 2.000.750000000000 ... 300.000.020942212934 To improve the convergence of the summation process note that 1 1 1 ( 1) 1 + = − k k + k k (2) which implies (1) =1.0. One can then produce a series for (x) – (1) which converges faster than the original series. This series not only converges much faster, it also reduces roundoff loss. This process of finding a faster converging series may be repeated again on the second series to produce a third sequence, which converges even more rapidly than the second. The following equation is helpful in determining how may items are required in summing the series above. for 1 and 1 1 1 1  >     −  = dx k r k n x r k n r