<s>[INST] Finding the value of $\sum_{n=1}^\infty\frac{n^2}{(n+1)(n+2)(n+3)(n+4)}$ Problem_

Find the value of $$\sum_{n=1}^\infty\frac{n^2}{(n+1)(n+2)(n+3)(n+4)}$$

It seems like I have to use the partial sum in order to get the exact value. But making it into the partial fractions was not that easy to me because of the numerator.
I tried to segregate the fraction by putting $A, B, C$ and $D$ as:
$$\frac{n^2}{(n+1)(n+2)(n+3)(n+4)}={A\over n+1}+{B\over n+2}+{C\over n+3}+{D\over n+4}$$
By multiplying both hands with $(n+1)(n+2)(n+3)(n+4)$, I could get the values which are $A=1/6, B=-2, C=9/2$ and $D=-8/3$. Unfortunately, there were no significant properties among them. 
I've also thought of dividing the numerator into two or more terms such as
$$n^2=n(n+1) - n$$
However, this method also did not give me any hint to solve. Are there other ways to evaluate the series looks like that? Thank you for your interest.
 [/INST] First add and subtract 1 to the numerator
it becomes $\sum_{n=1}^\infty \frac{(n+4-5)}{(n+2)(n+3)(n+4)} + \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)(n+3)(n+4)}$
$=\sum_{n=1}^\infty\frac{1}{(n+2)(n+3)} -\sum_{n=1}^\infty\frac{5}{(n+2)(n+3)(n+4)}+\sum_{n=1}^\infty\frac{1}{(n+1)(n+2)(n+3)(n+4)}$.(try evaluating this yourself)
$=\frac{1}{3}-\frac{5}{2} \times \frac{1}{12} + \frac{1}{3} \times \frac{1}{24} = \boxed{\frac{5}{36}}$
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