Solve $\sum_{n=1}^p\frac{2^ne^{nx}}{\binom{2n}n}$

Question

Related to another question, I need to solve the following summation:

$$\sum_{n=1}^p\frac{2^ne^{nx}}{\binom{2n}n}$$

Solved in terms of $x$ and $p$, and $\binom{2n}n=\frac{(2n)!}{(n!)^2}$

I could not make any progress...

If it helps, I want to take this to $p\to\infty$.

By 'solve', do you mean find a closed form? – sbares May 14 '16 at 11:23 — sbares, May 14 '16 at 11:23
@SBareS Yes please – Simply Beautiful Art May 14 '16 at 11:27 — Simply Beautiful Art, May 14 '16 at 11:27

score 2 · Accepted Answer · answered Jun 30 '17 at 16:56

Mathematica seems to think we have $$ \sum_{n=1}^p\frac{2^ne^{nx}}{\binom{2n}n}=\frac{1}{2 e^{-x}-1}+\frac{2 e^{x/2} \sin ^{-1}\left(\frac{e^{x/2}}{\sqrt{2}}\right)}{\left(2-e^x\right)^{3/2}}-\frac{2^{p+1} e^{p x+x} \, _2F_1\left(1,p+2;p+\frac{3}{2};\frac{e^x}{2}\right)}{\binom{2 (p+1)}{p+1}} $$ this looks much nicer with $2y=e^x$ and $q-1=p$ as $$ \sum_{n=1}^{q-1}\frac{4^ny^n}{\binom{2n}n}=\frac{y}{1-y}+\frac{\sqrt{y} \sin ^{-1}\left(\sqrt{y}\right)}{(1-y)^{3/2}}-\frac{4^q y^q \, _2F_1\left(1,q+1;q+\frac{1}{2};y\right)}{\binom{2 q}{q}} $$ we see only the right most terms depends on $q$, so it is likely that the sum telescopes to some extent. As $q\to \infty$, that term vanishes.

I should've expected lol, that binomial leads to pochhammer which obviously leads to hypergeometric series. Duh! — Simply Beautiful Art, Jun 30 '17 at 17:18

Solve $\sum_{n=1}^p\frac{2^ne^{nx}}{\binom{2n}n}$

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