show that $P(X=n) = \frac{4}{n(n+1)(n+2)}$ is pmf, that is show that$\sum^{\infty}_{n=1} P(X = n) = 1$ show that $P(X=n) = \frac{4}{n(n+1)(n+2)}$ is a pmf, that is show that$\sum^{\infty}_{n=1} P(X = n) = 1$
My solution 
$\sum^{\infty}_{n=1} P(X = n)$
$ =\sum^{\infty}_{n=1} \frac{4}{n(n+1)(n+2)}$
$ =\sum^{\infty}_{n=1} \frac{1}{n(n+1)} - \frac{1}{n(n+2)}$
$ = 2\sum^{\infty}_{n=1} (\frac{1}{n} - \frac{1}{(n+1)}) - (\frac{1}{n} - \frac{1}{(n+2)})$
$ = 2\sum^{\infty}_{n=1}  - \frac{1}{(n+1)} + \frac{1}{(n+2)}$
$ = -1$
What am I doing wrong here? 

Note that $$\frac{4}{n(n+1)(n+2)} = \frac 1 2\frac 1n-\frac 1{n+1}  +\frac 12\frac 1{n+2}$$
Hence your partial sums are $$\frac 1 2H_n-(H_{n+1}-H_1)+\frac 1 2 (H_{n+2}-H_2)$$
Using that $H_n=\log n+\gamma+O(1/n)$ you get the result. This generalizes to $$\sum_{n=1}^\infty \frac{k!}{(n)_k}=\frac{1}{k!k}$$
