Find $\sum_{n=1}^{\infty}$ $\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$ 
$$\sum_{n=1}^{\infty}\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$$

MY Approach$\sum_{n=1}^{\infty}$$\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$
= Lim$_{k\rightarrow\infty}$$\sum_{k=1}^{n}\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$
$\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$=
$\frac{1}{6\left(n+1\right)}$-$\frac{2}{\left(n+2\right)}$+$\frac{9}{\left(n+3\right)2}$-$\frac{8}{\left(n+4\right)3}$
I don't think i can telescope it, And i don't know any other method

Another approach , applicable to this type of sums in general, is to decompose the fraction as a sum
of fractions with Rising Factorials at the denominator.
$$
\eqalign{
  & {{n^{\,2} } \over {\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right)}} = {{\left( {n + 1} \right)\left( {n + 2} \right) - 3n - 2} \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} =   \cr 
  &  = {{\left( {n + 1} \right)^{\,\overline {\,2\,} }  - 3\left( {n + 1} \right) + 1} \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} = {1 \over {\left( {n + 3} \right)^{\,\overline {\,2\,} } }} - 3{1 \over {\left( {n + 2} \right)^{\,\overline {\,3\,} } }} + {1 \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} \cr} 
$$
That is of advantage because:
 - a polynomial at the numerator (here the $n^2$) can always be expressed (through Stirling Numbers) into a "polynomial" in rising/falling "powers";
 - the sum of Falling factorials parallels the integral of normal powers;
 - you can easily express the sum also between finite bounds.
We have in fact this equivalence between Falling and Rising Factorials
$$
x^{\underline {\, - q\,} }  = {1 \over {\left( {x + q} \right)^{\underline {\,q\,} } }} = {1 \over {\left( {x + 1} \right)^{\overline {\,q\,} } }}
$$
and the indefinite summation of the Rising Factorial is
$$
\sum\nolimits_x {{1 \over {\left( {x + 1} \right)^{\overline {\,q\,} } }}}  = \sum\nolimits_x {x^{\underline {\, - q\,} } }  = {1 \over {1 - q}}x^{\underline {\,1 - q\,} }  + c =  - {1 \over {\left( {q - 1} \right)\left( {x + 1} \right)^{\overline {\,q - 1\,} } }} + c
$$
as it easy to check by taking the Forward Finite Difference of both sides.
Therefore
$$
\sum\limits_{n = 1}^\infty  {{1 \over {\left( {n + m} \right)^{\overline {\,q\,} } }}} \quad \left| {\;2 \le q} \right.\quad  = {1 \over {\left( {q - 1} \right)\left( {m + 1} \right)^{\overline {\,q - 1\,} } }}
$$
and the result is
$$
\eqalign{
  & \sum\limits_{n = 1}^\infty  {{{n^{\,2} } \over {\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right)}}}  =   \cr 
  &  = \sum\limits_{n = 1}^\infty  {\left( {{1 \over {\left( {n + 3} \right)^{\,\overline {\,2\,} } }} - 3{1 \over {\left( {n + 2} \right)^{\,\overline {\,3\,} } }} + {1 \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }}} \right)}   \cr 
  &  = {1 \over {1\left( {3 + 1} \right)^{\overline {\,1\,} } }} - {3 \over {2\left( {2 + 1} \right)^{\overline {\,2\,} } }} + {1 \over {3\left( {1 + 1} \right)^{\overline {\,3\,} } }} =   \cr 
  &  = {1 \over 4} - {1 \over 8} + {1 \over {72}} = {{5} \over {36}} \cr} 
$$
For more details, also refer to this and to this other related post.
