<s>[INST] 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? 
 [/INST] The second step should be 
$$\sum_{n=1}^\infty \frac{2}{n(n+1)}- \frac{2}{(n+1)(n+2)},$$
and then the sum telescopes so we have $2(\frac{1}{2})=1$. The one half comes from the first part of the first term of $\sum_{n=1}^\infty \frac{1}{n(n+1)}- \frac{1}{(n+1)(n+2)}.$
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