<s>[INST] How to increase speed of convergence of series? I'm trying to speed up convergence of series:
$$
a_{n} = \frac{n(n-0.3)}{(n^{2} + 0.3)^{2}}
$$
Without series speed up:
$$
R_{method} = \left|A - S_{N_{0}}\right| = \sum_{n = N_{0} + 1}^{\infty}\frac{n(n-0.3)}{(n^{2} + 0.3)^{2}} \leq \sum_{n = N_{0} + 1}^{\infty}\frac{n^2}{n^{4}} < \int_{N_{0}}^{\infty}\frac{1}{x^2}dx = \frac{1}{N_{0}} < 0.25 \times 10^{7} \rightarrow N_{0} > 4 \times 10^{7}
$$
$$
S_{0} = \sum_{n = 1}^{4\times 10^{7}}\frac{n(n-0.3)}{(n^{2} + 0.3)^{2}} \approx 0.959433..
$$
$$
R_{calc} = N_{0}\times A_{a_{n}} < 0.25 \times 10^{-7} \rightarrow A_{a_{n}} < 0.625 \times 10^{-15}
$$
First speed up:
$$
\alpha_{0} = \lim_{n \rightarrow \infty}\frac{a_{n}}{\frac{1}{n^{2}}} = \lim_{x \rightarrow \infty } = \lim_{n \rightarrow \infty} \frac{(n - 0.3)n^{3}}{(n^2 + 0.3)^{2}} = 1
$$
$$
b_{n} =  \frac{n(n-0.3)}{(n^{2} + 0.3)^{2}} - \frac{1}{n^{2}} = \frac{-30n^{3}-60n^{2} -9 }{n^{2}(10n^2 + 3)^2}
$$
$$
B = \sum_{n = 1}^{\infty}b_{n}
$$
$$
R_{method} = \left|B - S_{N_{1}} \right| =\left| \sum_{n = N_{1} + 1}^{\infty} \frac{30n^3+60n^2+9}{n^2(10n^2+3)^2}\right| \leq \left| \sum_{n = N_{1} + 1}^{\infty}\frac{30n^3+60n^2+9}{10n^6}\right| <
$$
$$   
< \int_{N_{1}}^{\infty}\frac{30n^3+60n^2+9}{10n^6} = \frac{25(3N_{1}+4)N_{1}^2+ 9}{25N_{1}^5} < 0.25 \times 10^{-7} \rightarrow N_{1} > 10995.1
$$
$$
S_{1} = \sum_{n = 1}^{10996}\frac{-30n^{3}-60n^{2} -9 }{n^{2}(10n^2 + 3)^2} + \frac{\pi^2}{6} \approx 18021 
$$

There is a mistake but I cannot see it. Where did I go wrong and how should I have done it right?

 [/INST] Your error seems to lie in this computation:
$$\int_{N_{1}}^{\infty}\frac{30n^3+60n^2+9}{10n^6}dn= \frac{25(3N_{1}+4)N_{1}^2+ 9}{25N_{1}^5}$$
Indeed a primitive function of the integrand is:
$$-\frac32 \frac{1}{n^2}-\frac12 \frac{1}{n^3}-\frac{9}{10} \frac{1}{n^5}=- \frac{75n^3+25n^2+9}{50 n^5}$$
</s>