$$\sum_{k=1}^{\infty} \frac{H_{k}^{(2)}}{2^kk} = \frac{5\zeta(3)}{8}$$
Can someone show this without making a generating function for
$$\sum_{k=1}^{\infty} \frac{x^kH_{k}^{(2)}}{k}$$
Since that is the route I took and successfully got the answer. I totally feel as if there’s a nicer way to go about it, as the answer is so simple in terms of the zeta function.
To be clear: $$H_{n}^{(m)} = \sum_{k=1}^{n} \frac{1}{k^m}$$
The closed form follows on plugging $x=1/2$ in the generating function
$$\small{\sum_{n=1}^\infty\frac{H_{n}^{(2)}}{n}x^{n}=\operatorname{Li}_3(x)+2\operatorname{Li}_3(1-x)-\ln(1-x)\operatorname{Li}_2(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)}$$
note that $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^2(2)$ and $\operatorname{Li}_3(1/2)=\frac78\zeta(3)-\frac12\ln(2)\zeta(2)+\frac16\ln^3(2)$
Different approach
Recall the generalization
$$\int_0^1 \frac{\ln^a(1-x)\ln(1+x)}{x}dx=(-1)^a a! \sum_{n=1}^\infty\frac{H_n^{(a+1)}}{n2^n}$$
With $a=1$ we have
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n2^n}=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx$$ which is trivial.