Why is $\frac{1}{m+1} = \sum_{n=0}^\infty \frac{2^{-2n-m-1} (2n+m)!}{(n+m+1)!n!}$?

Question

I have the strong assumption that for $m \in \mathbb N_0$ we have $$\frac{1}{m+1} = \sum_{n=0}^\infty \frac{2^{-2n-m-1} (2n+m)!}{(n+m+1)!n!}=\sum_{n=0}^\infty \frac{2^{-2n-m-1}}{2n+m+1}\binom{2n+m+1}{n}.$$ But I didn't find an explicit way to show it.

In case you wanna know where this series is from: I'm still trying to solve How to prove an equality involving an infinity series and the modified Bessel functions?. If you plug in $s=\alpha=\beta=1$ you eventually come up with the series equation from above. At least the series equation would show the special case (in fact the other direction is not true).

Answer 1

A pssible idea could be to use

$$f(x)=\sum_{n=0}^\infty \frac{2^{-(2n+m+1)}\, (2n+m)!}{(n+m+1)!\,\,n!}x^{-n}=\frac{ \, _2F_1\left(\frac{m+1}{2},\frac{m+2}{2};m+2;\frac 1x\right)}{(m+1)\,\,2^{(m+1)}}$$ which is true even if $m$ is not an integer.

Now, working a little the expression

$$f(x)=\frac x {m+1} \left(1+\sqrt{\frac{x-1}{x}}\right)^{-m}\left(1-\sqrt{\frac{x-1}{x}}\right)$$

Answer 2

I guess I post my approch which is based on the idea of Claude Leibovici for clarity.

Using Legendre duplication we have $$(2n+m)!=\Gamma(2n+m+1) = \Gamma(n+ \tfrac{m+1}{2})\Gamma(n+ \tfrac{m+2}{2})2^{2n+m}/\sqrt{\pi}.$$

So we have $$\sum_{n=0}^\infty \frac{2^{-2n-m-1} (2n+m)!}{(n+m+1)!n!}=\frac{1}{2\sqrt{\pi}} \sum_{n=0}^\infty \frac{ \Gamma(n+ \tfrac{m+1}{2})\Gamma(n+ \tfrac{m+2}{2})}{\Gamma(n+m+2)n!}\\ = \frac{1}{2\sqrt{\pi}} \frac{ \Gamma(\tfrac{m+1}{2})\Gamma(\tfrac{m+2}{2})}{\Gamma(m+2)} F(\tfrac{m+1}{2},\tfrac{m+2}{2},m+2,1)\\ \frac{1}{2\sqrt{\pi}} \frac{ \Gamma(\tfrac{m+1}{2})\Gamma(\tfrac{m+2}{2})}{\Gamma(m+2)} \frac{\Gamma(m+2)\Gamma(m+2-\tfrac{m+1}{2}-\tfrac{m+2}{2})}{\Gamma(m+2-\tfrac{m+1}{2})\Gamma(m+2-\tfrac{m+2}{2})}\\ = \frac{1}{2\sqrt{\pi}} \frac{ \Gamma(\tfrac{m+1}{2})\Gamma(\tfrac{m+2}{2})}{\Gamma(m+2)} F(\tfrac{m+1}{2},\tfrac{m+2}{2},m+2,1)\\ \frac{1}{2\sqrt{\pi}} \frac{ \Gamma(\tfrac{m+1}{2})\Gamma(\tfrac{m+2}{2})}{\Gamma(m+2)} \frac{\Gamma(m+2)\Gamma(\tfrac{1}{2})}{\Gamma(\tfrac{m}{2}+\tfrac{3}{2})\Gamma(\tfrac{m}{2}+\tfrac{2}{2})} =\frac{1}{m-1}. $$