Is it correct that the series ($n,m>0$) $$ \sum_{k=0}^\infty\binom{2k+n}{k}\frac{1}{4^k\sqrt{k+m}} $$ diverges as $\sim\frac{2^n}{\sqrt{k(k+m)}}$?
If so what is the value of $$ \lim_{K\rightarrow\infty}\frac{1}{\log K}\sum_{k=0}^K\binom{2k+n}{k}\frac{1}{2^{2k+n}\sqrt{k}}? $$
By Stirling's approximation, $$ \binom{2\,k}{k}\sim\frac{1}{\sqrt{\pi\,k}}\,4^k. $$ Since $$ \binom{2\,k+n}{k}\ge\binom{2\,k}{k}, $$ the series diverges for all $n,m$.