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

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 $$\mathcal{S} = \lim_{p\to\infty} \sum_{n=1}^p \frac{n^2}{p_4(n)} = \lim_{p\to\infty}( g(1) - g(p+1)) = g(1) = \frac{6+9+5}{6\cdot 2\cdot 3 \cdot 4} = \frac{5}{36}$$

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

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,

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

$$=\frac{5}{36}$$

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.