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


$$\sum_{n \geq 1} \frac{(n!)(n^2)}{(n+4)!}$$


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,
$$=\frac{1}{6} \int_{0}^{1} x(x+1) dx$$
$$=\frac{5}{36}$$
