I have been wondering if anyone knows if there is a generating function for harmonic series of the form $H_{2n}$?.
That is, we are familiar with $$-\frac{\log(1-x)}{1-x}=\sum_{n=1}^{\infty}H_{n}x^{n}$$
But, is there one for $$\sum_{n=1}^{\infty}H_{2n}x^{n}$$ that anyone knows of?.
On
Yes. Take the even part and replace $x^2 \mapsto x$.
Edit: As an addendum to that glib answer, note that similar tricks can be used to get every $n$th term of a generating function $H(x)$: Compute the sum $$\frac{1}{n}\sum_{k=1}^n H(e^{2\pi i k/n})=\sum_{j=0}^\infty H_j x^j \left(\frac{1}{n}\sum_{k=1}^n e^{2\pi i j k/n }\right)$$ But this last sum of principle roots must be invariant under multiplication by $e^{2\pi i k/n}$ and so vanishes unless $n|j$ (and equals one otherwise). So only every $n$th term starting at $j=0$ survives into the generating function. (To start it at $j=s$ instead, just throw an extra factor of $e^{-2\pi i s/n}$ into the summation.)
It is an ugly trick, but it works. Given that: $$f(z)=\sum_{n=1}^{+\infty}H_n z^n = -\frac{\log(1-z)}{1-z},$$ then: $$\frac{f(z)+f(-z)}{2}=\sum_{n=1}^{+\infty}H_{2n} z^{2n},$$ hence: $$\sum_{n=1}^{+\infty} H_{2n} z^n = -\frac{1}{2}\left(\frac{\log(1-\sqrt{z})}{1-\sqrt{z}}+\frac{\log(1+\sqrt{z})}{1+\sqrt{z}}\right).$$