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
 A: This is one way to rewrite the summand into a form suitable for telescoping by hand.
Let $p_k(n) = \prod_{\ell=1}^k (n+\ell)$. The series at hand has the form
$$\mathcal{S} \stackrel{def}{=} \sum_{n=1}^\infty \frac{n^2}{(n+1)(n+2)(n+3)(n+4)} =
\sum_{n=1}^\infty \frac{n^2}{p_4(n)}$$
Notice we have the identity
$$\frac{1}{p_{k+1}(n)} = \frac{1}{k} \frac{(n+k+1)-(n+1)}{p_{k+1}(n)} = \frac{1}{kp_k(n)} - \frac{1}{kp_k(n+1)}$$
Repeat apply this identity to the summand of above series, we find
$$\begin{align}
\frac{n^2}{p_4(n)} &= \frac{n^2}{3p_3(n)} - \frac{n^2}{3p_3(n+1)}
= \color{red}{\frac{2n-1}{3p_3(n)}} + \color{green}{\frac{(n-1)^2}{3p_3(n)}} - \frac{n^2}{3p_3(n+1)}\\
\color{red}{\frac{2n-1}{3p_3(n)}} &= \frac{2n-1}{6p_2(n)} - \frac{2n-1}{6p_2(n+1)}
= \color{orange}{\frac{1}{3p_2(n)}} + \color{blue}{\frac{2n-3}{6p_2(n)}} - \frac{2n-1}{6p_2(n+1)}\\
\color{orange}{\frac{1}{3p_2(n)}} &= \color{magenta}{\frac{1}{3p_1(n)}} - \frac{1}{3p_1(n+1)}
\end{align}$$
Combine these, we can rewrite the summand as $\displaystyle\;\frac{n^2}{p_4(n)} = g(n) - g(n+1)$
where
$$\begin{align}
g(n) 
&=\color{magenta}{\frac{1}{3p_1(n)}} + \color{blue}{\frac{2n-3}{6p_2(n)}} + \color{green}{\frac{(n-1)^2}{3p_3(n)}}\\
&= \frac{\color{magenta}{2(n+2)(n+3)} + \color{blue}{(2n-3)(n+3)} + \color{green}{2(n-1)^2}}{6(n+1)(n+2)(n+3)}\\
&= \frac{6n^2+9n+5}{6(n+1)(n+2)(n+3)}
\end{align}$$
The series is now a telescoping one and
$$\mathcal{S} = \lim_{p\to\infty} \sum_{n=1}^p \frac{n^2}{p_4(n)} = \lim_{p\to\infty}( g(1) - g(p+1)) = g(1) = \frac{6+9+5}{6\cdot 2\cdot 3 \cdot 4} = \frac{5}{36}$$
A: 
$$\sum_{n \geq 1} \frac{(n!)(n^2)}{(n+4)!}$$


Note,
$$a_n=\frac{1}{6} \frac{(n!)(3!)}{(n+4)!}=\frac{1}{6}\frac{\Gamma(n+1) \Gamma (4)}{\Gamma (n+5)}$$
$$=\frac{1}{6}B(n+1,4)$$
$$=\frac{1}{6} \int_{0}^{1}x^n (1-x)^3 dx$$
Where $\Gamma$ denotes the Gamma Function, and $B$ denotes the Beta Function. We have utilized the Beta-Gamma Function relationship.

We are interested in:
$$\sum_{n \geq 1} n^2a_n$$
$$=\frac{1}{6} \int_{0}^{1} \left(\sum_{n \geq 1} n^2 x^n \right) (1-x)^3 dx$$

It is standard to show,
$$\sum_{n \geq 1} n^2 x^n=\frac{x(x+1)}{(1-x)^3}$$
By first considering $\sum_{n \geq 1} x^n=\frac{1}{1-x}$ and (repeatedly) differentiating then multiplying by $x$ on both sides. Convergence of the above series is for $|x|<1$ by the ratio test.

Hence we get,
$$=\frac{1}{6} \int_{0}^{1} x(x+1) dx$$
$$=\frac{5}{36}$$
A: The partial fractions should come out to be $$\frac 16\cdot\frac 1{r+1}-2\cdot\frac 1{r+2}+\frac 92\cdot \frac 1{r+3}-\frac 83\cdot\frac 1{r+4}$$
Examine now what happens to the fractions with denominator $5$, which comes with $r=1,2,3,4$ and you get $$\frac 15\cdot\left(-\frac 83+\frac 92-2+\frac 16\right)=0$$So you do get telescoping. I'll leave you to fill in the remaining details.
A: 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.
