Genereating function of $H_{2n}$ 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. 
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\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}
A: 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.
