<s>[INST] value of an $\sum_3^\infty\frac{3n-4}{(n-2)(n-1)n}$ I ran into this sum $$\sum_{n=3}^{\infty} \frac{3n-4}{n(n-1)(n-2)}$$
I tried to derive it from a standard sequence using integration and derivatives, but couldn't find a proper function to describe it.
Any ideas?
 [/INST] Setting 
$$\frac{3n-4}{n(n-1)(n-2)}=\frac{A(n-1)-B}{(n-2)(n-1)}-\frac{An-B}{(n-1)n}$$
gives you $A=3,B=2$, i.e.
$$\frac{3n-4}{n(n-1)(n-2)}=\frac{3(n-1)-2}{(n-2)(n-1)}-\frac{3n-2}{(n-1)n}.$$
Hence, we have
$$\begin{align}\sum_{n=3}^{\infty}\frac{3n-4}{n(n-1)(n-2)}&=\lim_{m\to\infty}\sum_{n=3}^{m}\left(\frac{3(n-1)-2}{(n-2)(n-1)}-\frac{3n-2}{(n-1)n}\right)\\&=\lim_{m\to\infty}\left(\frac{3\cdot 2-2}{1\cdot 2}-\frac{3m-2}{(m-1)m}\right)\\&=2\end{align}$$
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