<s>[INST] Calculate the sum $\sum_{n=3}^{\infty}\frac{4n-3}{n^3-4n}$ $$\sum_{n=3}^{\infty}\frac{4n-3}{n^3-4n}$$
I think it is related to power series, because it is the topic, but I have no idea how to get there.
Could you give a hint?
 [/INST] By partial fractions
$$\begin{align}\sum\limits_{n=3}^\infty\frac{4n-3}{n^3-4n}&=\sum\limits_{n=3}^\infty\frac{5}{8(n-2)}+\frac{3}{4n}-\frac{11}{8(n+2)}
\\&=\sum\limits_{n=1}^4\frac{5}{8n}+\sum\limits_{n=3}^4\frac{3}{4n}+\sum\limits_{n=5}^\infty\frac{5}{8n}+\frac{3}{4n}-\frac{11}{8n}
\\&=\sum\limits_{n=1}^4\frac{5}{8n}+\sum\limits_{n=3}^4\frac{3}{4n}
\\&=\frac{167}{96}
\end{align}$$
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