Generalized Harmonic Number Summation $ \sum_{n=1}^{\infty} {2^{-n}}{(H_{n}^{(2)})^2}$ 
Prove That $$ \sum_{n=1}^{\infty} \dfrac{(H_{n}^{(2)})^2}{2^n} = \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 $$
Notation : $ \displaystyle H_{n}^{(2)} = \sum_{r=1}^{n} \dfrac{1}{r^2}$

We can solve the above problem using the generating function $\displaystyle \sum_{n=1}^{\infty} (H_{n}^{(2)})^2 x^n $, but it gets rather tedious especially taking into account the indefinite polylogarithm integrals involved. Can we solve it using other methods like Euler Series Transform or properties of summation?

second approach suggested by Cornel Ioan Valean using summation by parts and lets start with the following sum:
with ${N \in \mathbb{N}_{\ \geq\ 1}}$
\begin{align}
\sum_{n=1}^N\frac{\left(H_{n-1}^{(2)}\right)^2}{2^n}=\sum_{n=1}^N\frac{\left(H_n^{(2)}\right)^2}{2^n}-2\sum_{n=1}^N\frac{H_n^{(2)}}{n^22^n}+\sum_{n=1}^N\frac1{n^42^n}\tag{1}
\end{align}
on the other hand:
\begin{align}
\sum_{n=1}^N\frac{\left(H_{n-1}^{(2)}\right)^2}{2^n}=\sum_{n=1}^{N-1}\frac{\left(H_{n}^{(2)}\right)^2}{2^{n+1}}=\sum_{n=1}^{N}\frac{\left(H_{n}^{(2)}\right)^2}{2^{n+1}}-\frac{\left(H_{N}^{(2)}\right)^2}{2^{N+1}}\tag{2}
\end{align}
from $(1)$ and $(2)$ we reach
$$\sum_{n=1}^N\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=4\sum_{n=1}^N\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^N\frac{1}{n^42^n}-2\frac{\left(H_{N}^{(2)}\right)^2}{2^{N+1}}$$
letting $N$ approach $\infty$ we get
$$\sum_{n=1}^\infty\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=4\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{1}{n^42^n}-0$$
I was able here to prove $$\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42
\end{align*}$$
which follows $$\sum_{n=1}^\infty\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=2\operatorname{Li_4}\left(\frac{1}{2}\right)+\frac14\zeta(4)+\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42
$$
