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?
 A: By partial fractions
$$\begin{align}\sum\limits_{n=3}^\infty\frac{4n-3}{n^3-4n}&=\sum\limits_{n=3}^\infty\frac{5}{8(n-2)}+\frac{3}{4n}-\frac{11}{8(n+2)}
\\&=\sum\limits_{n=1}^4\frac{5}{8n}+\sum\limits_{n=3}^4\frac{3}{4n}+\sum\limits_{n=5}^\infty\frac{5}{8n}+\frac{3}{4n}-\frac{11}{8n}
\\&=\sum\limits_{n=1}^4\frac{5}{8n}+\sum\limits_{n=3}^4\frac{3}{4n}
\\&=\frac{167}{96}
\end{align}$$
A: This link shows you the partial fraction decomposition.
If you carefully look at the terms cancelling out, you should be able 
to calculate the sum.
A: see if this helps.
$$\frac{a}{(n-2)n} - \frac{b}{n(n+2)}  = \frac{(a-b)n+2(a+b)}{(n-2)n(n+2)}$$  choose $a, b$ so that $$ a- b = 4, 2a + 2b = -3 \to a = \frac54, b = -\frac{11}4$$
A: You can write your sum as the sum of four telescoping sums by noting that
\begin{eqnarray*}
\frac{4n-3}{n^{3}-4n} &=&\frac{3/4}{n}+\frac{5/8}{\left( n-2\right) }-\frac{%
11/8}{\left( n+2\right) } \\
&=&\left( \frac{5/8}{n-2}-\frac{5/8}{n-1}\right) +\left( \frac{5/8}{n-1}-%
\frac{5/8}{n}\right) +\left( \frac{11/8}{n}-\frac{11/8}{n+1}\right) +\left( 
\frac{11/8}{n+1}-\frac{11/8}{n+2}\right) 
\end{eqnarray*}
can you take it from here? 
