Improving the convergence of the series

Question

I am trying to calculate the sum of a A = $\sum_{1}^{\infty}\frac{1}{n^2 + 0.4}$ with precision $0.5\cdot 10^{-7}$

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$R_{method} = |A - S_{N}| = \sum_{N+1}^{\infty}\frac{1}{n^2 + 0.4}$

$\sum_{N+1}^{\infty}\frac{1}{n^2 + 0.4} < \int_{N}^{\infty}\frac{dx}{x^2 + 0.4} < \int_{N}^{\infty}\frac{dx}{x^2} = \frac{1}{N} < 0.25 \cdot 10^{-7} \rightarrow N > 4 \cdot 10^7$

$S = \sum_{1}^{10^7}\frac{1}{n^2 + 0.4} = 1.3288275$

I have improved the convergence:

$\lim_{n -> \infty} \frac{a_{n}}{\frac{1}{n^2}} = lim_{n -> \infty}\frac{n^2}{0.4 + n^2} = 1$

$b_{n} = \frac{1}{n^2 + 0.4} - \frac{1}{n^2} = \frac {-0.4}{(n^2 + 0.4)n^2}$

$B = \sum b_{n} $

$R_{method} = |B - \overline{S_{N_0}}| = \sum_{N_{0} + 1}^{\infty}\frac {0.4}{(n^2 + 0.4)n^2} < \int_{N_{0}}^{\infty}\frac{0.4 dx}{x^4} = \frac{2} {15 N_{1}^3} < 0.25*10^{-7} \rightarrow N_{0} > 200 \cdot (\frac{2}{3})^\frac{1}{3}$

$200 \cdot (\frac{2}{3})^\frac{1}{3} < 175 $

$S_{0} = \sum_{1}^{175}\frac{-0.4}{(n^2 + 0.4)n^2} + \frac{\pi^2}{6} = 1.3288276$

Please tell me: Why aren't 7 digits after the decimal point equal in $S_{0}\: and \: S$?

Answer 1

Let's take some more digits and terms. For the second approach

 N        S2[N]
 173   1.3288276216043133
 174   1.3288276211679408
 175   1.3288276207414575
 176   1.3288276203245846
 177   1.3288276199170528
2000   1.3288275960922136

so that this result appears to be correctly rounded.

In the first, simple, approach, using the correct bound $\frac1N\le 0.5\cdot 10^{-7}\implies N\ge 2\cdot 10^7$, I get the partial sum at $N=2\cdot 10^7$ as $S_{1,N}=1.3288275460737846$, which is within the required distance to the more correct result, even if it rounds to a different final digit. This is normal, there is no way to get always the desired number of correct digits be prescribing absolute or relative error tolerances.

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It is also useful to use the bounds of the remainder of the integral test to improve the accuracy. For instance, you found that $$ \int_{N+1}^\infty\frac1{x^2+0.4}dx\le \sum_{n=N+1}^\infty\frac1{n^2+0.4}\le\int_{N}^\infty\frac1{x^2+0.4}dx \\ \sqrt{2.5}\arctan\frac{\sqrt{0.4}}{N+1}\le \sum_{n=N+1}^\infty\frac1{n^2+0.4}\le\sqrt{2.5}\arctan\frac{\sqrt{0.4}}{N} $$ This would improve the first summation to $$ S_N+\sqrt{2.5}\arctan\frac{\sqrt{0.4}}{N+1}\le A\le S_N+\sqrt{2.5}\arctan\frac{\sqrt{0.4}}{N} \\ 1.328827596073782 \le A\le 1.3288275960737845 $$ which is much more accurate than desired. The difference between the bounds is essentially $\frac1{N^2}$, so using $N>\sqrt{2\cdot 10^7}=\sqrt{20}\cdot 10^3$ gives $N=5000$ as a nice number, where the above bounds result in $$ 1.328827576082216 \le A\le 1.3288276160742167 $$ which indeed has the required accuracy and not more.