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!
 A: Hint
Do a change of index to express all the sums on the form $\sum_n \frac 1 n$ and simplify.
Edit First notice as user127001 said you are not allowed to split the sum on three divergent sum so the key is to use a partial sum and finaly you pass to the limit.
Now from your work
$$\frac{3}{4}\sum_{n=3}^N{ \frac{1}{n}} + \frac{5}{8}\sum_{n=3}^N \frac{1}{n-2} - \frac{11}{8}\sum_{n=3}^N \frac{1}{n+2}=\frac{3}{4}\sum_{n=3}^N{ \frac{1}{n}}+\frac{5}{8}\sum_{n=1}^{N-2} \frac{1}{n}- \frac{11}{8}\sum_{n=5}^{N+2} \frac{1}{n}$$
and we cancel all the terms from $n=5$ to $n=N-2$ and we pass to the limit $N\to\infty$ to conclude.
A: $$\sum_{3}^{\infty} \frac{ 4n-3 }{ n^3-4n }=\\
\sum_{3}^{\infty} \frac{ 4n-3 }{ n(n^2-4) }=\\
\sum_{3}^{\infty} \frac{ 4n }{ n(n-2)(n+2) }+\sum_{3}^{\infty} \frac{ -3 }{ n(n-2)(n+2) }=\\
4\sum_{3}^{\infty} \frac{ 1 }{ (n-2)(n+2) }-3\sum_{3}^{\infty} \frac{ +1 }{ n(n-2)(n+2)}$$
A: 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 $n \to \infty$.
