<s>[INST] How to find what the sum of this infinite series is. I have the following series:
$\sum_{n\geq3} \frac{4n-3}{n^3-4n}$
which I've simplified into the following:
$\frac{3}{4}\sum_{n\geq3}{ \frac{1}{n}} + \frac{5}{8}\sum_{n\geq3} \frac{1}{n-2} - \frac{11}{8}\sum_{n\geq3} \frac{1}{n+2}$ 
And this is where I'm stuck... How do I calculate the total sum?
Thanks in advance!
 [/INST] Just a simple generalization: denote $S_n=\sum_{j=3}^{n}\frac{1}{j}$. Then your whole expression becomes (after partial fraction expansion):
$$
V=\lim_{n \to \infty}\bigg(\frac{3}{4}S_n+\frac{5}{8}\bigg(S_n+\frac{1}{2} +1 +\varphi_1(n)\bigg)-\frac{11}{8}\bigg(S_n-\frac{1}{3}-\frac{1}{4} +\varphi_2(n)\bigg)\bigg)=\frac{5}{8} \cdot \frac{3}{2}+\frac{11}{8}\cdot \frac{1}{12}
$$
because $\varphi_1(n)$ and $\varphi_2(n) \ \to 0$ as $n \to \infty$.
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