We know the generating function of: $$\sum_{n=1}^{\infty}H_nx^n=\frac{\ln(1-x)}{x-1}$$.
How do we find out the generating function of $$\sum_{n=1}^{\infty}H_{2n}x^n$$
I used the formula: $\displaystyle { H }_{ 2n }=\frac { 1 }{ 2 } \left[ { H }_{ n }+{ H }_{ n-\frac { 1 }{ 2 } } \right] +\ln { 2 } $. But that didn't help.
On
Define $$f(x) = \sum_{n=1}^\infty H_{2n} x^{2n}, \quad g(x) = \sum_{n=1}^\infty H_{2n-1} x^{2n-1}.$$ Then since $$H_{2n} = H_{2n-1} + \frac{1}{2n},$$ we have $$f(x) = \sum_{n=1}^\infty H_{2n} x^{2n} = x \sum_{n=1}^\infty H_{2n-1} x^{2n-1} + \sum_{n=1}^\infty \frac{x^{2n}}{2n} = x g(x) - \frac{1}{2}\log(1-x^2).$$ But since $$f(x) + g(x) = \sum_{m=1}^\infty H_m x^m = \frac{\log(1-x)}{x-1} = h(x),$$ we have $$f(x) = x (h(x) - f(x)) - \frac{1}{2} \log(1-x^2),$$ or $$f(x) = \frac{1}{1+x} \left( x h(x) - \frac{\log (1-x^2)}{2} \right).$$ Then the desired sum is simply $$f(x^{1/2}) = \frac{2 x^{1/2} \tanh^{-1} x^{1/2} - \log(1-x)}{2(1-x)} $$ which is a straightforward algebraic exercise.
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}H_{2n}x^{n}} & = \sum_{n = 1}^{\infty}H_{2n}x^{2n/2} = \sum_{n = 1}^{\infty}H_{n}x^{n/2}\,{1 + \pars{-1}^{n} \over 2} \\[3mm] & = \half\sum_{n = 1}^{\infty}H_{n}\pars{\root{x}}^{n} + \half\sum_{n = 1}^{\infty}H_{n}\pars{-\root{x}}^{n} \\[3mm] & = \color{#f00}{-\,\half\bracks{{\ln\pars{1 - \root{x}} \over 1 - \root{x}} + {\ln\pars{1 + \root{x}} \over 1 + \root{x}}}} \end{align}