How to reduce (or simplify) a power series

Question

So I have the following power series from my book and I am trying to reduce it. I became a bit confused about how to reduce it and was wondering how you would go about simplifying it.

Answer 1

Start with the known generating function of the central binomial coefficients: $$\sum_{k=0}^\infty \binom{2 k}{k} z^k = \frac{1}{\sqrt{1 - 4 z}}.$$ Differentiate with respect to $z$ and multiply by $z$: $$\sum_{k=0}^\infty k \binom{2 k}{k} z^k = \frac{2z}{(1 - 4 z)^{3/2}}.$$

Now the original sum is \begin{align} \sum_{k=0}^\infty \frac{(2k+1)!}{2^{3k+1}k!^2} &=\frac{1}{2}\sum_{k=0}^\infty (2k+1) \binom{2 k}{k} \left(\frac{1}{8}\right)^k \\ &= \sum_{k=0}^\infty k\binom{2 k}{k} \left(\frac{1}{8}\right)^k +\frac{1}{2}\sum_{k=0}^\infty \binom{2 k}{k} \left(\frac{1}{8}\right)^k\\ &= \frac{2(1/8)}{(1 - 4(1/8))^{3/2}} + \frac{1}{2}\cdot\frac{1}{\sqrt{1 - 4(1/8)}}\\ &= \frac{1/4}{(1/2)^{3/2}} + \frac{1}{2\sqrt{1/2}}\\ &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\\ &=\sqrt{2}. \end{align}