Solve Summation of Catalan Convolution? 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:
$$4\sum_{k\geq 0}{2k+3\choose k}\frac{\left(\frac 1 2\right)^{2k+4}}{k+4}=1$$
