How find this series $\sum_{n=1}^{\infty}\frac{1}{n^2H_{n}}$? 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
 A: While finding closed form is not likely, there's a way to get a good numerical value for this slowly converging series.
Turns out, there exist very sharp inequalities for harmonic numbers, found in this paper: https://www.sciencedirect.com/science/article/pii/S0723086906000168.

$$a-\ln \left(e^{1/(n+1)}-1 \right) \leq H_n < b-\ln \left(e^{1/(n+1)}-1 \right)$$

Where:
$$a=1+\ln \left(\sqrt{e}-1 \right)$$
$$b=\gamma$$
The equality case on the left is only for $n=1$.
The inequalitites can be used to estimate the tail of the series after computing several terms. Let's denote:
$$S_N=\sum_{n=1}^N \frac{1}{n^2 H_n}$$
$$T_N=\sum_{n=N+1}^\infty \frac{1}{n^2 H_n}$$
$$f(n)=\ln \left(e^{1/(n+1)}-1 \right)$$
We can write:
$$T_N<T_{aN}=\sum_{n=N+1}^\infty \frac{1}{n^2 \left(a-f(n)\right)}$$
$$T_N>T_{bN}=\sum_{n=N+1}^\infty \frac{1}{n^2 \left(b-f(n)\right)}$$
The sums on the right can be again estimated using integrals:
$$\int_N^\infty \frac{dx}{x^2~\left(a-f(x)\right)}<T_{aN}<\int_{N+1}^\infty \frac{dx}{x^2~\left(a-f(x)\right)}$$
The integrals likely don't have a closed form, however they can be evaluated numerically with good precision (after a suitable change of variable):
$$\int_N^\infty \frac{dx}{x^2~\left(a-f(x)\right)}=\int_0^{1/(N+1)} \frac{dy}{(1-y)^2 \left(a-\ln (e^y-1)\right)}$$

For example, we can take $N=100$, then we have only three correct digits by truncating the series:
$$S_{100}=\color{blue}{1.33}1114\dots$$
On the other hand, using the method above we can estimate the tail:
$$T_{a100}=0.00165\dots$$
$$T_{b100}=0.00163\dots$$
$$0.00163\dots< T_{100} < 0.00165\dots$$
So we obtain:

$$\color{blue}{1.3327}4\dots<S_{\infty}<\color{blue}{1.3327}6\dots$$

We have obtained $5$ correct digits using just $100$ terms. Taking the arithmetic mean of the two bounds, we get $6$ correct digits.

I might improve the numerical results later with Mathematica (for this answer I only used Wolfram Alpha).
