Compute $\sum\limits_{k=0}^{100}\frac{1}{(100-k)!(100+k)!}$ $$\sum_{k=0}^{100}\frac{1}{(100-k)!(100+k)!}$$
My work 
$$\sum_{k=0}^{100}\frac{2n!}{(2n!)(n-k)!(n+k)!}$$
$$\sum_{k=0}^{100}\frac{^{2n}C_{n-k}}{(2n!)}$$
$$\sum_{k=0}^{100}\frac{^{2n}C_{n+k}}{(2n!)}$$
How I should proceed further ?
 A: Probably too complex for an answer.
Considering that $$\sum_{k=0}^{n}\frac{x^k}{(n-k)!(n+k)!}=\frac{\, _2F_1(1,-n;n+1;-x)}{(n!)^2}$$ Setting $x=1$, then $$\sum_{k=0}^{n}\frac{1}{(n-k)!(n+k)!}=\frac{\Gamma \left(n+\frac{1}{2}\right)+\sqrt{\pi } \,\Gamma (n+1)}{2 (n!)^2 \Gamma
   \left(n+\frac{1}{2}\right)}=\frac{1}{2} \left(\frac{1}{\Gamma (n+1)^2}+\frac{4^n}{\Gamma (2 n+1)}\right)=\frac{1}{2} \left(\frac{1}{(n!)^2}+\frac{4^n}{(2 n)!}\right)$$ Using Stirling approximation for $k!$, an approximation could be $$\frac{e^{2 n} \left(\sqrt{\pi n} +1\right)}{4 \pi n^{(2 n+1)} }$$
A: Solving the problem for $n \in \mathbb{N}$ instead of only $n = 100$, you get
\begin{align}
\sum_{k=0}^{n} \frac{1}{(n-k)!(n+k)!}
&= \frac{1}{(2n)!} \sum_{k=n}^{2n} \frac{(2n)!}{k!(2n-k)!}
= \frac{1}{(2n)!} \sum_{k=n}^{2n} \binom{2n}{k}  \\
&= \frac{1}{2(2n)!} \left( \binom{2n}{n} + 2^{2n} \right) \\
\end{align}
In the last step you use the fact that
$$
\sum_{k=n}^{2n} \binom{2n}{k} = \frac{1}{2} \left( \binom{2n}{n} + 2^{2n} \right)
$$
which is shown by
\begin{align}
2^{2n} 
&= \sum_{k=0}^{2n} \binom{2n}{k} \\
&= \sum_{k=0}^{n-1} \binom{2n}{k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k}  \\
&= \sum_{k=n+1}^{2n} \binom{2n}{2n-k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k}  \\
&= \sum_{k=n+1}^{2n} \binom{2n}{k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k}  \\
&= 2 \sum_{k=n}^{2n} \binom{2n}{k} ~-~ \binom{2n}{n}
\end{align}
from which you get
$$
\frac{1}{2} \left( 2^{2n} + \binom{2n}{n} \right)
= \sum_{k=n}^{2n} \binom{2n}{2n-k} 
$$
A: Trying to parse your request, we want to compute (for $n=100$, or maybe for any $n\in\mathbb{N}$)
$$ \sum_{k=0}^{n}\frac{1}{(n-k)!(n+k)!} = \sum_{k=0}^{n}\frac{1}{k!(2n-k)!} = [x^{2n}]\left(\sum_{k=0}^{n}\frac{x^k}{k!}\right)^2=\frac{1}{(2n)!}\sum_{k=0}^{n}\binom{2n}{k}. $$
Due to the symmetry of binomial coefficients,
$$ \sum_{k=0}^{n}\binom{2n}{k} = \frac{1}{2}\left( 4^n+\binom{2n}{n}\right),$$
hence:

$$ \sum_{k=0}^{n}\frac{1}{(n-k)!(n+k)!} = \color{red}{\frac{4^n}{2\cdot(2n)!}+\frac{1}{2\cdot n!^2}}.$$

