Recently I tried to estimate this sum but without much success, would you have any methods to suggest?
$S=\sum_{n=1}^{+\infty} (-1)^n \frac{H_{2n}}{n^2}$
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1Since you want to estimate it, and is an alternating sum with decreasing terms, the sum is between any two consecutive partial sums. – marty cohen Jul 20 '23 at 18:46
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Also, Wolfram Alpha says sum_(n=1)^∞ ((-1)^n sum_(k=1)^(2 n) 1/k)/n^2≈-1.14961134098672142140504700316342630968969743214600308575545207339825767678712015922092651572286220525471 – marty cohen Jul 20 '23 at 18:56
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How did you try to estimate it? Also what else have you tried? – Varun Vejalla Jul 20 '23 at 19:09
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There is almost certainly a duplicate of this sum on this site. – user170231 Jul 20 '23 at 19:11
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5Does this answer your question? How to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}$. Found using approach0.xyz – Varun Vejalla Jul 20 '23 at 19:13
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Thanks @VarunVejalla the closest I could find was a solution for $\sum (-1)^n \dfrac{\color{red}{H_n}H_{2n}}{n^2}$ – user170231 Jul 20 '23 at 19:15
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thanks you so much @VarunVejalla !!! – azur Jul 21 '23 at 05:36