TITLE: How find this series $\sum_{n=1}^{\infty}\frac{1}{n^2H_{n}}$?
QUESTION [25 upvotes]: Question:

This follow series have simple closed form?
  $$\sum_{n=1}^{\infty}\dfrac{1}{n^2H_{n}}$$
  where 
  $$H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$

I suddenly thought of this series, because I know this series 
$$\dfrac{1}{n^2H_{n}}\approx\dfrac{1}{n^2\ln{n}}$$
since
$$\lim_{n\to\infty}\dfrac{\dfrac{1}{n^2\ln{n}}}{\dfrac{1}{n^2}}=0$$
so
$$\sum_{n=2}^{\infty}\dfrac{1}{n^2\ln{n}}$$  converges. So
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2H_{n}}$$ converges. But this sum have closed form? such as
link  sum
and 
Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$
 Thank you

REPLY [2 votes]: This approach does not lead to closed form, but allows to obtain the required sum with the arbitrary accuracy.
$\color{brown}{\textbf{Preliminary notes}}$
Firstly, the harmonic sum can be presented via digamma function,
$$H_n = \gamma+\psi_0(n+1),\tag1$$
then
$$S=\lim\limits_{N\to\infty} S(N),\tag2$$
where
$$S(N) = \sum_{n=1}^N\frac1{n^2(\gamma+\psi_0(n+1))}.\tag3$$
Secondly, the sum $S(N)$ reaches its limit slowly.
For example, $S(100000)\approx1.33275'48120$ and $S(2000000)\approx1.33275'55487.$
Thirdly, are known the asymptotic series
$$\psi_0(1+z) = \log z - \frac1{2z} - \sum_{m=1}^\infty\frac{B_{2m}}{2mz^{2m}}.\tag4$$
This gives the motivation to estimate $S-S(N).$
$\color{brown}{\textbf{Estimaton}}$
Using two first terms of the sum in $(4),$ one can get
$$n^2H_n = n^2(\log n+ \gamma) - \frac{60n^3+10n^2-1}{120n^4}+\dots = n^2(\log n+ \gamma)\left(1 - \frac{60n^3+10n^2-1}{120n^6(\log n+ \gamma)}+\dots\right),$$
$$S-S(N)=\sum_{n=N+1}^\infty\frac1{n^2(\log n + \gamma)}\sum_{j=0}^\infty\left( \frac{60n^3+10n^2-1}{120n^6(\log n + \gamma)}+\dots\right)^j,$$
$$S-S(N)\approx\sum_{n=N+1}^\infty\sum_{j=1}^\infty \frac{(60n^3+10n^2-1)^{j-1}}{120^{j-1}n^{6j-4}(\log n + \gamma)^j},\tag5$$
$$S\approx \sum_{n=1}^N\frac1{n^2(\gamma+\psi_0(n+1))} + \sum_{n=N+1}^\infty\sum_{j=1}^\infty \frac{(60n^3+10n^2-1)^{j-1}}{120^{j-1}n^{6j-4}(\log n + \gamma)^j}.\tag6$$
Calculations of formulas $(3),(5),(6)$ give:
\begin{pmatrix}
N & S(N) & S-S(N) & S\\
10 & 1.30752'69686 & 0.02542'72 & 1.33295'42\\
20 & 1.32179'08559 & 0.01100'11 & 1.33279'20\\
50 & 1.32906'32996 & 0.00369'629 & 1.33275'959\\
100 & 1.33111'40166 & 0.00164'234 & 1.33275'546\\
200 & 1.33201'79513 & 0.00073'7784 & 1.33275'5735\\
500 & 1.33249'56726 & 0.00025'9926 & 1.33275'5599\\
1000 & 1.33263'64101 & 0.00011'9174 & 1.33275'5584\\
2000 & 1.33270'05664'81413 & 0.00005'50143 & 1.33275'55807\\
5000 & 1.33273'56014'83833 & 0.00001'99785 & 1.33275'55800\\
10000 & 1.33274'62412'69296 & 0.00000'93386'5 & 1.33275'55799'2\\
20000 & 1.33275'11959'85027 & 0.00000'43839'2 & 1.33275'55799'1\\
50000 & 1.33275'39573'66156 & 0.00001'16225'3 & 1.33275'55799'0\\
100000 & 1.33275'48120'15120 & 0.00000'07678'3 & 1.33275'55799'0\\
200000 & 1.33275'52154'37624 & 0.00000'03644'57 & 1.33275'55798'95\\
500000 & 1.33275'54432'72882 & 0.00000'01866'22 & 1.33275'55798'95\\
1000000 & 1.33275'55146'82723 & 0.00000'00692'173 & 1.33275'55798'950\\
\end{pmatrix}
$\color{nrown}{\mathbf{Conclusions}}$

$S\approx1.33275557895.$.
Formula $(6)$ has increased accuracy compared to direct calculations.
The accuracy of formula $(6)$ can be increased to the required by using additional summands in the sum of $(4)$.
To reduce the formula $(6)$ to a closed form, we need at least a closed form for sums of the form
$$\sum_{n=N+1}^\infty\frac1{n^p(\gamma+\log n)^q}.$$