<s>[INST] How to prove $\sum _{k=1}^{\infty } (-1)^k H_{\frac{2 k}{3}} = -\frac{\pi }{2 \sqrt{3}}+\frac{3 \pi }{8}-\frac{3}{4} \log (2)$? I stumbled on this problem in the wake of the discussion  https://math.stackexchange.com/a/3553902/198592
Can you make sense of the equation in the question involving the harmonic number with a rational index 
$H_{\frac{2}{3} k}$ although the series is not convergent?
Hint: find the generating function of $g(z) = \sum_{k=1}^{\infty}z^k H_{\frac{2 k}{3}}$ and interpret the sum as the limit $z\to -1$.
 [/INST] Start with using the integral representation of the harmonic number $H_n=\int_0^1\frac{1-x^n}{1-x}\ dx$ we have 
$$\sum_{k=0}^\infty(-1)^k H_{\frac{2k}{3}}=\int_0^1\frac{1}{1-x}\sum_{k=0}^\infty((-1)^k-(-x^{\frac23})^n)\ dx$$
$$\int_0^1\frac{1}{1-x}\left(\frac12-\frac{1}{1+x^{\frac23}}\right)\ dx\overset{x\to x^3}{=}-\frac32\int_0^1\left(\frac{x}{1+x^2}-\frac{1}{1+x^2}+\frac{1}{1+x+x^2}\right)\ dx$$
$$=-\frac32\left[\frac12\ln(1+x^2)-\tan^{-1}x+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{1+2x}{\sqrt{3}}\right)\right]_0^1$$
$$= -\frac{3}{4} \ln2+\frac{3 \pi }{8}-\frac{\pi }{2 \sqrt{3}}$$
Note that I used Grandi series $\sum_{k=0}^\infty (-1)^k=\frac12$.  
</s>