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.
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By the generalized binomial theorem,$$(1+4z)^{-1/2}=\sum_{n\ge0}\frac{(-1/2)_n}{n!}4^nz^n=\sum_{n\ge0}\binom{2n}{n}(-z)^n.$$Differentiating,$$-2(1+4z)^{-3/2}=\sum_{n\ge0}\binom{2n}{n}(-1)^nnz^{n-1}=\sum_{n\ge1}\binom{2n}{n}(-1)^nnz^{n-1}.$$Multiplying by $z$,$$-2z(1+4z)^{-3/2}=\sum_{n\ge1}\binom{2n}{n}n(-z)^n.$$
J.G.
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$$n\binom{2n}n=n\cdot\dfrac{(2n!)}{n! n!}=2\binom{2n-1}{n-1}$$
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 $m=-\dfrac32, x=4z$
lab bhattacharjee
- 274,582
$$\displaystyle\sum_{n=0}^\infty \binom {2n}nx^n=\frac1{\sqrt{1-4x}}$$
which is a consequence of the (generalized) binomial theorem, and converges for $|x|<\frac14$.
– player3236 Oct 22 '20 at 11:18