<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] 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.
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