Evaluate $\sum_{k=1}^n\frac{(2k)!!}{(2k-1)!!}z^k$

Question

I want to find a closed form for the following:
$$J_k=\int_0^1\frac{x^k}{x+1}\operatorname{arsinh}^2(\sqrt{x}/2)dx$$Using the power series for these 2 functions we get$$=\int_0^1x^k\left(\sum_{n=0}^\infty(-1)^nx^n\right)\left(\frac{1}{2}\sum_{m=0}^\infty\frac{(-1)^{m}(2x)^{2m+2}}{(m+1)^2{{2m+2}\choose{m+1}}}\right)$$
Using the Cauchy Product we get:$$=\frac{1}{2}\int_0^1x^k\sum_{m=0}^\infty\sum_{n=0}^m(-1)^{m-n}x^{m-n}\frac{(-1)^n4^{n+1}x^{2n+2}}{(n+1)^2{2n+2\choose n+2}}dx$$I simplified and switched the order of the sums and the integral. Evaluating that integral we get$$=\frac{1}{2}\sum_{m=0}^\infty(-1)^{m}\sum_{n=0}^m\frac{4^{n+1}}{(n+1)^2(m+n+k+3){2n+1\choose n+1}}$$I wanted to get rid of the $(m+n+k+3)$, so I set up an integral inside the inner sum$$=\frac{1}{2}\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}\sum_{n=0}^m\frac{4^{m+n+k+3}}{(n+1)^2(m+n+k+3){2n+2\choose n+1}}$$$$=\frac{1}{2}\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}\sum_{n=0}^m\frac{1}{(n+1)^2{2n+2\choose n+1}}\int_0^4x^{m+n+k+2}dx$$Then I took the integral to the outside$$=\frac{1}{2}\int_0^4\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}\sum_{n=0}^m\frac{x^{m+n+k+2}}{(n+1)^2{2n+2\choose n+2}}dx$$
My best idea after this was to make another integral
$$=\frac{1}{2}\int_0^4\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}x^{m+k+1}\sum_{n=0}^m\frac{x^{n+1}}{(n+1)(n+1){2n+2\choose n+2}}dx$$$$=\frac{1}{2}\int_0^4\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}x^{m+k+1}\sum_{n=0}^m\frac{1}{(n+1){2n+2\choose n+2}}\int_0^xy^ndy\space dx$$$$=\frac{1}{2}\int_0^4\int_0^x\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}x^{m+k+1}\sum_{n=0}^m\frac{y^n}{(n+1){2n+2\choose n+2}}dy\space dx$$
My approach from here was to make another integral
$$=\frac{1}{2}\int_0^4\int_0^x\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}\frac{x^{m+k+1}}{y}\sum_{n=0}^m\frac{1}{{2n+2\choose n+2}}\int_0^yz^n dz\space dy\space dx$$$$=\frac{1}{2}\int_0^4\int_0^x\int_0^y\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}\frac{x^{m+k+1}}{y}\sum_{n=0}^m\frac{z^n}{{2n+2\choose n+2}} dz\space dy\space dx$$
Reindex
$$=\frac{1}{2}\int_0^4\int_0^x\int_0^y\sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}\frac{x^{m+k+1}}{yz}\sum_{n=1}^{m+1}\frac{z^n}{{2n\choose n}} dz\space dy\space dx$$Using
$${2n\choose n}=4^n\frac{(2n-1)!!}{(2n)!!}$$
We get$$\sum_{n=1}^{m+1}\frac{z^n}{{2n\choose n}}=\sum_{n=1}^{m+1}\frac{(z/4)^n(2n)!!}{{(2n-1)!!}}$$
We can reduce the problem to finding $$\sum_{k=1}^n\frac{(2k)!!}{(2k-1)!!}z^k$$
I am not sure how to go from here. I don't know any integral representation for $\frac{(2k)!!}{(2k-1)!!}$, and I don't have too much experience in sums

Answer 1

Starting from $$ S= \sum_{m=0}^\infty(-1)^{m}4^{-m-k-2}\frac{x^{m+k+1}}{yz}\sum_{n=1}^{m+1}\frac{z^n}{\binom{2n}{n}} $$ then \begin{align} S &= \sum_{m=0}^\infty(-1)^{m}4^{-m-k-2} \, \frac{x^{m+k+1}}{yz} \, \sum_{n=0}^{m} \frac{z^{n+1}}{\binom{2n+2}{n+1}} \\ &= \frac{x^{k+1}}{4^{k+2} \, y} \, \sum_{m=0}^{\infty} \left(\frac{-x}{4}\right)^{m} \, \sum_{n=0}^{m} \frac{z^{n}}{\binom{2n+2}{n+1}} \\ &= \frac{x^{k+1}}{4^{k+1} \, y \, (x+4)} \, \sum_{n=0}^{\infty} \frac{(-1)^n}{\binom{2n+2}{n+1}} \, \left( \frac{x \, z}{4} \right)^n. \end{align} Now, $$ \frac{1}{\binom{2n}{n}} = \frac{n! \, n!}{(2n)!} = \frac{\Gamma^{2}(n+1)}{\Gamma(2n+1)} = (2n+1) \, B(n+1, n+1) $$ which leads to $$ \frac{1}{\binom{2n}{n}} = (2n+1) \, \int_{0}^{1} t^{n} \, (1-t)^{n} \, dt $$ and \begin{align} S_{1} &= \sum_{n=0}^{\infty} \frac{x^n}{\binom{2n+2}{n+1}} \\ &= \int_{0}^{1} \, t (1-t) \, \sum_{n=0}^{\infty} (2n+3) \, (x \, t \, (1-t))^n \, dt \\ &= \int_{0}^{1} \left(\frac{t (1-t)}{1 - x t (1-t)} + \frac{2 t (1-t)}{(1 - x \, t \, (1-t))^2} \right) \, dt \end{align} Using \begin{align} I_{1} &= \int_{0}^{1} \frac{t (1-t) \, dt}{1 - x t (1-t)} \\ &= \frac{1}{x} \, \left[-t + \frac{2}{\sqrt{x(4-x)}} \, \tan^{-1}\left( \frac{(2 t -1) \, \sqrt{x}}{\sqrt{4-x}} \right) \right]_{0}^{1} \\ &= \frac{1}{x \, \sqrt{x (4-x)}} \, \left(-\sqrt{x (4-x)} + 4 \, \tan^{-1}\left(\sqrt{\frac{x}{4-x}}\right) \right) \\ I_{2} &= \int_{0}^{1} \frac{t (1-t) \, dt}{(1 - x t (1-t))^2} \\ &= \frac{1}{x \, \sqrt{x(4-x)}} \, \left[\frac{(1-2t) \, \sqrt{x}}{1 - x t (1-t)} - \frac{2 (x-2)}{\sqrt{4-x}} \, \tan^{-1}\left(\frac{(2t-1) \, \sqrt{x}}{\sqrt{4-x}} \right) \right]_{0}^{1} \\ &= \frac{1}{x \, \sqrt{x(4-x)}} \, \left(-2 \sqrt{x} - \frac{4 (x-2)}{\sqrt{4-x}} \, \tan^{-1}\left(\sqrt{\frac{x}{4-x}}\right) \right) \end{align} then \begin{align} S_{1} &= \frac{\phi(x)}{x^2} \, \left(-4 \, \sqrt{x} - \frac{x}{\phi(x)} + 4 \, \left(1 - \frac{2(x-2)}{\sqrt{4-x}}\right) \, \tan^{1}(\phi(x)) \right), \end{align} where $\phi(x) = \sqrt{\dfrac{x}{4-x}}$.

This leads to $$ S = \frac{x^{k}}{4^{k} \, y \, z \, (x+4) \, \phi} \, \left(8 \, \sqrt{x z} + \frac{x z}{\phi} + 16 \, \left(1 + \frac{x z + 8}{\sqrt{x z + 16}} \right) \, \tanh^{-1}(\phi) \right), $$ where $\phi = \sqrt{\dfrac{x z}{x z + 16}}$.

In this scheme the result presented here doesn't seem to lead to anything efficient enough to reduce the calculations desired.