Simplify $\sum_{k=0}^{n} \frac{1}{(n-k)!(n+k)!}$

Question

I have to simplify the following sum ($n \in \mathbb{N}$). I have tried to give values to $n$, but I haven't noticed anything useful. Can you help me, please? Thanks! $$\sum_{k=0}^{n} \frac{1}{(n-k)!(n+k)!}$$

Answer 1

$$\sum_{k=0}^{n} \frac{1}{(n-k)!(n+k)!}$$

$$= \frac {1}{n!n!} + \frac {1}{(n-1)!(n+1)!} + ... + \frac {0!}{2n!}$$

$$= \frac {1}{(2n)!} \bigg( \binom {2n}{n} +\binom {2n}{n-1} +\binom {2n}{n-2}+ ... + \binom {2n}{0}\bigg)$$

$$= \frac {1}{(2n)!} \sum _{k=0}^n \binom {2n}{k}=\frac {1}{2(2n)!} \bigg (4^n+ \binom {2n}{n}\bigg )$$

Answer 2

Hints:

It equals $$\frac12\left(\sum_{i=0}^{2n}\frac1{i!(2n-i)!}+\frac1{n!n!}\right)$$

Now multiply by $(2n)!$, calculate and divide the result by $(2n)!$

Answer 3

$$\frac{1}{(2n)!}\sum_{k=0}^{n}\binom{2n}{n+k} = \frac{1}{2(2n)!}\left[\binom{2n}{n}+\sum_{k=0}^{2n}\binom{2n}{k}\right]=\frac{1}{2\cdot n!^2}+\frac{4^n}{2\cdot(2n)!}.$$