$\sum _ { n = 1 } ^ { \infty } n \binom {2 n} {n}( - z )^n$ According to wolframalpha,
$$\sum _ { n = 1 } ^ { \infty } n \binom {2 n} {n} ( - z )^n =  - \frac { 2 z } { ( 4 z + 1 ) ^ { 3 / 2 } } \text { when } 4 | z | < 1$$ However I have been unable to find how to get this result.

$$n\binom{2n}n=n\cdot\dfrac{(2n!)}{n! n!}=2\binom{2n-1}{n-1}$$
Now like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $.
comparing the expansion of $(1+x)^m$ with $$2\sum_{n=1}^\infty\binom{2n-1}{n-1} (-z)^{n-1}$$
$$mx=2\binom31(-z)^{2-1}\text{ and }\dfrac{m(m-1)}2x^2=2\binom52(-z)^{3-1}$$
to find $m=-\dfrac32, x=4z$
