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?
 A: Hint:
$$\frac{3n-4}{n(n-1)(n-2)} \equiv \frac{1}{n-1} + \frac{1}{n-2} - \frac{2}{n}$$ 
A: Using the Heaviside Method of Partial Fractions to solve
$$
\frac{3n-4}{n(n-1)(n-2)}=\frac{A}{n}+\frac{B}{n-1}+\frac{C}{n-2}
$$
we get
multiply by $n$ and set $n=0\implies\frac{3\cdot\color{#C00000}{0}-4}{(\color{#C00000}{0}-1)(\color{#C00000}{0}-2)}=A\implies A=-2$
multiply by $n-1$ and set $n=1\implies\frac{3\cdot\color{#C00000}{1}-4}{\color{#C00000}{1}(\color{#C00000}{1}-2)}=B\implies B=1$
multiply by $n-2$ and set $n=2\implies\frac{3\cdot\color{#C00000}{2}-4}{\color{#C00000}{2}(\color{#C00000}{2}-1)}=C\implies C=1$
Thus,
$$
\frac{3n-4}{n(n-1)(n-2)}=-\frac2n+\frac1{n-1}+\frac1{n-2}
$$
Then applying Telescoping Series to get
$$
\begin{align}
\sum_{n=3}^N\frac{3n-4}{n(n-1)(n-2)}
&=-2\sum_{n=3}^N\frac1n+\sum_{n=3}^N\frac1{n-1}+\sum_{n=3}^N\frac1{n-2}\\
&=\color{#C00000}{-2\sum_{n=3}^N\frac1n}\color{#00A000}{+\sum_{n=2}^{N-1}\frac1n}\color{#0000F0}{+\sum_{n=1}^{N-2}\frac1n}\\
&=\color{#0000F0}{1+\frac12}\color{#00A000}{+\frac12+\frac1{N-1}}\color{#C00000}{-\frac2{N-1}-\frac2N}
\end{align}
$$
Now take the proper limit.
Note that we can't sum from $n=0$ since the denominators will be $0$, so I assumed we were to start at $n=3$.
A: 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
$$\begin{align}\sum_{n=3}^{\infty}\frac{3n-4}{n(n-1)(n-2)}&=\lim_{m\to\infty}\sum_{n=3}^{m}\left(\frac{3(n-1)-2}{(n-2)(n-1)}-\frac{3n-2}{(n-1)n}\right)\\&=\lim_{m\to\infty}\left(\frac{3\cdot 2-2}{1\cdot 2}-\frac{3m-2}{(m-1)m}\right)\\&=2\end{align}$$
