Can someone help me solve this?
$$ \sum_{k=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{2 k+4} 4\left(\begin{array}{c} 2 k+3 \\ k \end{array}\right)}{k+4} $$ converges.
Wolfram alpha says it converges to $1$.
But actually, I have no idea, how to solve it.
Using a version of the binomial power series $$\sum_{k\geq 0}{2k+\alpha \choose k}z^k=\frac 1 {\sqrt{1-4z}}\left(\frac {1-\sqrt{1-4z}}{2z}\right)^\alpha$$ With $\alpha=3$: $$\sum_{k\geq 0}{2k+3\choose k}z^k=\frac 1 {\sqrt{1-4z}}\left(\frac {1-\sqrt{1-4z}}{2z}\right)^3$$ Consequently $$\sum_{k\geq 0}{2k+3\choose k}\frac{z^{k+4}}{k+4}=\int_0^z \frac {t^3} {\sqrt{1-4t}}\left(\frac {1-\sqrt{1-4t}}{2t}\right)^3dt=\frac{\left(1-\sqrt{1-4z}\right)^4}{64}$$ Evaluating at $z=\frac 1 4$: $$\sum_{k\geq 0}{2k+3\choose k}\frac{\left(\frac 1 2\right)^{2k+8}}{k+4}=\frac{1}{64}$$ Re-arranging the terms: $$4\sum_{k\geq 0}{2k+3\choose k}\frac{\left(\frac 1 2\right)^{2k+4}}{k+4}=1$$