<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] $$\sum_{3}^{\infty} \frac{ 4n-3 }{ n^3-4n }=\\
\sum_{3}^{\infty} \frac{ 4n-3 }{ n(n^2-4) }=\\
\sum_{3}^{\infty} \frac{ 4n }{ n(n-2)(n+2) }+\sum_{3}^{\infty} \frac{ -3 }{ n(n-2)(n+2) }=\\
4\sum_{3}^{\infty} \frac{ 1 }{ (n-2)(n+2) }-3\sum_{3}^{\infty} \frac{ +1 }{ n(n-2)(n+2)}$$
</s>