I've recently seen a nice binoharmonic series from Ali's book (page 309), but that post, for some reason, vanished (Update: that post was undeleted and now it may be found here Evaluating $\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x$).
It is an excellent addition to MSE, that is,
$$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_{2n}}{n^3}.$$
since finding elegant ways is not easy to do.
I would also add the following variant, which seems to be a new one (in case it is known, any reference will be highly appreciated), $$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_{2n}}{n^4}.$$ Can we find very elegant ways to go for these two series?
Lots of other variants we could consider, like $$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_{2n}^2}{n^2}.$$
More curious binoharmonic structures (only a few examples in the list of possible items alike, keeping the weight $\le5$), $$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_n H_{2n}}{n^2};$$ $$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_n H_{2n}}{n^3};$$ $$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_n H_{2n}}{(2n+1)^2};$$ $$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_n H_{2n}}{(2n+1)^3}.$$
Let me also make a little update (February 03, 2023):
$$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_n H_{2n}H_{4n}}{n^2};$$
$$ \sum_{n=1}^{\infty} \frac{4^n}{\displaystyle \binom{2n}{n}} \frac{H_n H_{2n}H_{4n}}{(2n+1)^2}.$$
Update: Yes, that's the answer to the question above concerning the first series, and we can get brilliant proofs, and a paper is under work, which will be available soon.
