It is clean harmonics

Goal: Evaluate

$\sum_{n=1}^{\infty} H_n^2 x^n$

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$S=\sum _{n=1}^{\infty }H_n^2x^n=\frac{1}{x}\sum _{n=2}^{ \infty}\left(H_n-\frac{1}{n}\right)^2x^n=\frac{1}{x}\left[\sum _{n=1}^{\infty}H_n^2x^n+\sum _{n=2}^{\infty}\frac{1}{n^2}x^n-2\sum _{n=2}^{\infty}\frac{H_n}{n}x^n\right]-1$

$=\frac{1}{x}\left[S+\text{Li}_2\left(x\right)-2\sum _{n=1}^{\infty}\frac{H_n}{n}x^n\right]$

$S=\left(\frac{1}{x-1}\right)\left(\text{Li}_2\left(x\right)-2\sum _{n=1}^{\infty}\frac{H_n}{n}x^n\right)$

To evaluate $\sum _{n=1}^{\infty}\frac{H_n}{n}x^n$ we can make use of the harmonic generating function $\sum _{n=1}^{ \infty}H_kx^k=\frac{\ln \left(1-z\right)}{z-1}$

Integrating the generating function gives

$\sum _{n=1}^{\infty}\frac{H_n}{n}x^n=\text{Li}_2\left(x\right)+\frac{1}{2}\ln ^2\left(1-x\right)$

Putting it all together gives

$S=\frac{\text{Li}_2\left(x\right)+\ln ^2\left(1-x\right)}{1-x}$

Substituting $x=1/2$ and using the reflection formula for the dilogarithm function gives

$S=\sum _{n=1}^{\infty }\frac{H_n^2}{2^n}=\frac{\pi ^2}{6}+\ln ^2\left(2\right)$

The answer follows