good question on summation

Question

$$\sum_{k=1}^{∞}\frac{(2k+1)\sum_{m=0}^{k-1} \frac{(2k+1)!}{(k+1-m)!(k-m-1)!(m+1)!m!}}{4^{2k+1}}$$ pls evaluate this. I tried to evaluate this but failed. Can anybody help me. I tried to find that if the summation was telescoping but failed. I also wrote a code This gives the upper summation as a convergent series. But how is this even possible. Any body please help me

Answer 1

$$\frac{(2k+1)!}{(k+1-m)!(k-m-1)!m!(m+1)!} = \binom{k+1}{m}\binom{k}{m+1}\binom{2k+1}{k}$$ so by Vandermonde's identity $$ \sum_{m=0}^{k-1}\frac{(2k+1)!}{(k+1-m)!(k-m-1)!m!(m+1)!}=\frac{4^{2k+1} k\, \Gamma\left(k+3/2\right)^2}{\pi\,\Gamma(k+2)\Gamma(k+3)} $$ and we want to evaluate $$ \frac{1}{\pi}\sum_{k\geq 1}\frac{(2k+1) k\, \Gamma\left(k+3/2\right)^2}{\Gamma(k+2)\Gamma(k+3)} $$ but this is not convergent, since $\frac{(2k+1) k\, \Gamma\left(k+3/2\right)^2}{\Gamma(k+2)\Gamma(k+3)}\to 2$ as $k\to +\infty$.