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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.

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  • $\begingroup$ You did not ask a question. $\endgroup$ Jan 30, 2023 at 16:21
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    $\begingroup$ @ Mariano Suárez-Álvarez He did, see line 9 of the OP. $\endgroup$ Jan 30, 2023 at 22:02
  • $\begingroup$ @Dr.WolfgangHintze, what is "a way to go"? (Line numbers do not mean anything here, btw) $\endgroup$ Jan 31, 2023 at 3:53
  • $\begingroup$ I deleted it because I remember I saw a similar post that I posted here on MSE before. I tried to find now but couldn't find it. I'll undelete it now $\endgroup$ Feb 2, 2023 at 3:41
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    $\begingroup$ No I have not but I think I can get the integral representations of these monster series and I think that's all I can do 😀 . Such series are quite challenging $\endgroup$ Feb 2, 2023 at 10:35

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