How to prove that $$\sum_{k=1}^\infty\frac1{k^2(k+1)}=\frac{\pi^2}6-1$$? I could do this by partial fractions and setting the summand equal to $$\frac{Ak(k+1)+B(k+1)+Ck^2}{k(k+1)}=\frac1{k(k+1)}$$But I want to see solutions other than that one.
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3Just show $\sum\limits_{k=1}^n\frac1{k^2(k+1)} =\sum\limits_{k=1}^n\frac1{k^2} + \frac{1}{n+1} - 1$ by induction and you end up with $\zeta(2)-1$ – Henry Mar 27 '23 at 16:58
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Kamal Saleh: I may have made a mistake, but I do not see where I should have $\frac{1}{k+1}$ - my intended brackets were $\left(\sum\limits_{k=1}^n\frac1{k^2}\right) + \frac{1}{n+1} - 1$ – Henry Mar 27 '23 at 17:06
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@Henry Oh nevermind, I messed up. Thanks! – Kamal Saleh Mar 27 '23 at 17:12
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In other words, $\frac1{k^2(k+1)}=\frac1{k^2}-\frac1{k(k+1)}$ and the sum of $\frac1{k(k+1)}$ can be calculated by telescoping: $\frac1{k(k+1)}=\frac1k-\frac1{k+1}$. – Kenta S Mar 27 '23 at 17:38
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1Why the downvotes? Please tell me what I did wrong so that I can edit accordingly. Thank you. – Kamal Saleh Mar 27 '23 at 18:21
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@KentaS Add this as an answer please :) – Kamal Saleh Mar 27 '23 at 18:22
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Does this answer your question? Sum of series with addition: $\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$ https://math.stackexchange.com/questions/1685733 – Anne Bauval Mar 31 '23 at 16:59
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After "setting the summand equal to...", there seems to be a square missing on $k$ at the denominators. – Anne Bauval Mar 31 '23 at 17:07
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Compute the partial sum$$S_n=\sum_{k=1}^n\frac1{k^2(k+1)}=\sum_{k=1}^n\frac1{k^2}+\sum_{k=1}^n\frac1{k+1}-\sum_{k=1}^n\frac1{k}$$ Using harmonic numbers $$S_n=H_n^{(2)}+\left(H_{n+1}-1\right)-H_n$$ Using their asymptotics $$S_n=\left(\frac{\pi ^2}{6}-1\right)-\frac{1}{2 n^2}+\frac{5}{6 n^3}+O\left(\frac{1}{n^4}\right)$$ which shows a relative error smaller than $0.001$% as soon as $n>19$
Claude Leibovici
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I wonder why the OP accepted that solution, after having asked for a solution "other than" "do this by partial fractions". Anyway, it was a (multi!)duplicate. https://approach0.xyz/search/?q=OR%20content%3A%24%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cfrac1%7Bk%5E2(k%2B1)%7D%24&p=1 – Anne Bauval Mar 31 '23 at 17:06