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. 

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.
