A Harmonic Sum

Let $S$ denote the summation.

The series is convergent by Ratio Test as :- $\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}=\frac{(H_{n+1})^{2}}{2(H_{n})^{2}}=\frac{1}{2}\\$ [ $a_{n}$ is the general nth term of the series]

Here we can use the approximation $H_{n} \approx \ln(n)+\gamma$ when $n\to\infty$ [Here $\gamma$ denotes the Euler-Mascheroni Constant]

$S = \frac{(H_{1})^{2}}{2^{1}} +\frac{(H_{2})^{2}}{2^{2}}+\frac{(H_{3})^{2}}{2^{3}}+\frac{(H_{4})^{2}}{2^{4}}+.......$

$\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\frac{S}{2}=\qquad\qquad\frac{(H_{1})^{2}}{2^{2}} +\frac{(H_{2})^{2}}{2^{3}}+\frac{(H_{3})^{2}}{2^{4}}+\frac{(H_{4})^{2}}{2^{5}}+.......$

Subtracting the second expression from the first expression.

$\frac{S}{2} = \sum_{n=1}^{\infty}\frac{\left((H_{n})^{2}-(H_{n-1})^{2}\right)}{2^{n}}$

$\displaystyle = \sum_{n=1}^{\infty}\frac{(H_{n}+H_{n-1})}{n2^{n}}$ [As $H_{n}-H_{n-1} = \frac{1}{n}$ ]

Now both the series $\displaystyle \sum_{n=1}^{\infty}\frac{H_{n}}{n2^{n}}$ and the series $\displaystyle \sum_{n=1}^{\infty}\frac{H_{n-1}}{n2^{n}}$ converges by ratio test.

Hence we can write $\displaystyle = \sum_{n=1}^{\infty}\frac{(H_{n}+H_{n-1})}{n2^{n}}=\sum_{n=1}^{\infty}\frac{H_{n}}{n2^{n}}+\sum_{n=1}^{\infty}\frac{H_{n-1}}{n2^{n}}=\sum_{n=1}^{\infty}\frac{H_{n}}{n2^{n}}+\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)2^{n+1}}$

Now using the Generating Function for $\displaystyle \sum_{n=1}^{\infty}H_{n}z^{n} = -\frac{\ln(1-z)}{1-z}$

We can write it in terms of two integrals:-

$\displaystyle \int_{0}^{\frac{1}{2}} \frac{-\ln(1-x)}{x(1-x)}dx - \int_{0}^{\frac{1}{2}}\frac{\ln(1-x)}{1-x}{dx} = -\int_{0}^{\frac{1}{2}}\left(\frac{\ln(1-x)}{x} +\frac{2\ln(1-x)}{1-x}\right)dx = \text{Li}_{2}(\frac{1}{2}) +\frac{(\ln(2))^{2}}{2} = \frac{\pi^{2}}{12} + \frac{(\ln(2))^{2}}{2}$

So $\displaystyle \frac{S}{2} =\frac{\pi^{2}}{12} + \frac{(\ln(2))^{2}}{2}$

So $\displaystyle S=\frac{\pi^{2}}{6} + (\ln(2))^{2}$

Note:- Here $\text{Li}_{2}(z)$ denotes the Dilogarithm Function .

I'm assuming that you are aware of the generating function of the harmonic numbers, IE, $\displaystyle\sum_{n=1}^{\infty} H_nz^n=-\dfrac{\log(1-z)}{1-z}$ . This is quite easy to prove by considering the recurrence relation of the harmonic numbers.

Integrate both the sides from $z=0$ to $z=x$ ( $0<x<1$ ) to obtain $\sum_{n=2}^{\infty}\dfrac{H_{n-1}}{n}x^n=\dfrac{1}{2}\log^2(x) \ \ \ \ \ \ \ \ \ \ \ ---- (1)$

Consider the following recurrence relation of the harmonic numbers $H_n=H_{n-1}+\dfrac{1}{n}$ Square both the sides to get $(H_n)^2=(H_{n-1})^2+\dfrac{1}{n^2}+2\dfrac{H_{n-1}}{n}$ Now let the generating function of the infinite sequence $\left\{\left(H_n\right)^2\right\}_{n=1}^{\infty}$ be $g(x)$ . Multiply both the sides of the equality by $x^n$ and sum it up from $n=2$ to $n=\infty$ to get $\begin{aligned}\sum_{n=2}^{\infty}\left(H_n\right)^2x^n &=\sum_{n=2}^{\infty} \left(H_{n-1}\right)^2x^n+\sum_{n=2}^{\infty} \dfrac{x^n}{n^2}+\sum_{n=2}^{\infty}\dfrac{H_{n-1}}{n}x^n\\g(x)-x &\stackrel{\text{from} \ (1)}{=}xg(x)+\rm{Li}_2(x)-x+\log^2(1-x) \\ g(x) &= \dfrac{\rm{Li}_2(x)+\log^2(1-x)}{1-x} \end{aligned}$ Now simply put $x=2$ to obtain $\displaystyle\sum_{n=1}^{\infty} \dfrac{\left(H_n\right)^2}{2^n}=g(1/2)=\boxed{\dfrac{\pi^2}{6}+\log^2(2)}$

NOTE 1 : $\rm{Li}_2(x)=\displaystyle\sum_{n=1}^{\infty} \dfrac{x^n}{n^2}$

NOTE 2 : To evaluate $\rm{Li}_2(1/2)$ , use the duplication formula - $\rm{Li}_2(x)+\rm{Li}_2(1-x)=\dfrac{\pi^2}{6}-\log(x)\log(1-x)$