How to prove this equation with Bessel function in summation How to prove this:
$$2\sum_{n=1}^{\infty}(-1)^n\frac{J_{2n}(z)}{n}=\sum_{m=1}^{\infty}\left\{\frac{(-1)^m(\frac{z}{2})^{2m}}{(m!)^2}\sum_{k=1}^{m}\frac{1}{k}\right\}$$
This equation occurs in converting standard BesselY function to Neumann's representation. 
You can find this problem in the book 'A Treatise on the Theory of Bessel Functions' by G. N. Watson, 1944, page 67. 
screenshot of page 67
Thank you very much.
 A: It is well-known that
$$ J_{2n}(z)=\sum_{l\geq 0}\frac{(-1)^l z^{2l+2n}}{2^{2l+2n}l!(2n+l)!} \tag{1}$$
and that $J_n(z)$ has a very fast decay to zero as $n\to +\infty$ for a fixed $z$. It follows that:
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{(-1)^n}{n}J_{2n}(z) &=& \sum_{n\geq 1}\sum_{l\geq 0}\frac{(-1)^{l+n}(z/2)^{2l+2n}}{nl!(2n+l)!}\\&=&\sum_{h\geq 1}(-1)^h (z/2)^{2h}\sum_{l=0}^{h-1}\frac{1}{(2h-l)!l!(h-l)}\end{eqnarray*}\tag{2} $$
and it is enough to find a closed form for the very last sum:
$$\begin{eqnarray*} \sum_{l=0}^{h-1}\frac{1}{(2h-l)!l!(h-l)} &=& \sum_{l=1}^{h}\frac{1}{l(h-l)!(h+l)!}\\&=&\color{blue}{\frac{1}{h!^2}}\sum_{l=1}^{h}\binom{h}{l}\frac{ h!(l-1)!}{(h+l)!}\end{eqnarray*} \tag{3}$$
Here comes the magic of Euler's Beta function:
$$\begin{eqnarray*} \sum_{l=1}^{h}\binom{h}{l}\frac{\Gamma(h+1)\,\Gamma(l)}{\Gamma(h+l+1)}&=&\int_{0}^{1}\sum_{l=1}^{h}\binom{h}{l}z^{l-1}(1-z)^h\,dz\\&=&\int_{0}^{1}\frac{(1+z)^h-1}{z}(1-z)^h\,dz\tag{4}\end{eqnarray*}$$
leading to:
$$\begin{eqnarray*} \sum_{l=1}^{h}\binom{h}{l}\frac{ h!(l-1)!}{(h+l)!} &=& \int_{0}^{1}\frac{(1-z^2)^h-1}{z}\,dz -\int_{0}^{1}\frac{(1-z)^h-1}{z}\,dz\\&=&\frac{1}{2}\int_{0}^{1}\frac{(1-z)^h-1}{z}\,dz-\int_{0}^{1}\frac{(1-z)^h-1}{z}\,dz\\&=&\frac{1}{2}\int_{0}^{1}\frac{1-z^h}{1-z}\,dz = \color{red}{\frac{H_h}{2}}\tag{5}\end{eqnarray*} $$
and finally to:

$$ \color{red}{2}\sum_{n\geq 1}\frac{(-1)^n}{n}J_{2n}(z) = \sum_{h\geq 1}\frac{(-1)^h (z/2)^{2h}}{\color{blue}{h!^2}}\color{red}{\sum_{k=1}^{h}\frac{1}{k}}.\tag{6}$$

