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Prove that $$ \frac{1}{1.2.3.4}+\frac{4}{3.4.5.6}+\frac{9}{5.6.7.8}+\frac{16}{7.8.9.10}+\dots=\frac{1}{6}\log2-\frac{1}{24} $$

My Attempt $$ T_n=\frac{n^2(2n-2)!}{(2n+2)!}=\frac{n^2}{(2n+2)(2n+1)(2n)(2n-1)}\\ =\frac{1}{24}\Big[\frac{-2}{n+1}+\frac{3}{2n+1}+\frac{2}{2n-1}\Big]\\ $$ $$ S=\frac{1}{24}\Big[{-2}\big[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots\big]+{3}\big[\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\big]+{2}\big[1+\frac{1}{3}+\frac{1}{5}+\dots\big]\Big]\\ =\frac{1}{24}\Big[{2}\big[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots\big]+{3}\big[\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\big]-{2}\big[\frac{1}{3}+\frac{1}{5}+\dots\big]\Big]\\ =\frac{1}{24}\Big[2\log2+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\Big]=\frac{1}{12}\log 2+\frac{1}{24}\Big[\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\Big] $$ I dont think that the last term series converges so I think I'm stuck, and what does it mean? How do I proceed further and evaluate the infinite series ?

Thanx to @Robert Z for the correction.

Method 1 $$ S=\frac{1}{24}\Big[-2[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots]+3[\frac{1}{3}+\frac{1}{5}+\dots]+[1+\frac{1}{3}+\frac{1}{5}+\dots]\Big]\\ =\frac{1}{24}\Big[\color{red}{-2\big[\frac{1}{3}+\frac{1}{5}+\dots\big]}+2\big[-\frac{1}{2}-\frac{1}{4}-\dots\big]+\color{red}{2\big[\frac{1}{3}+\frac{1}{5}+\dots\big]}+\big[\frac{1}{3}+\frac{1}{5}+\dots\big]+\big[1+\frac{1}{3}+\frac{1}{5}+\dots\big]\Big]\\ =\frac{1}{24}\Big[2(-\frac{1}{2}-\frac{1}{4}-\dots)+2(1+\frac{1}{3}+\frac{1}{5}+\dots)-1\Big]\\ =\frac{1}{24}[2\log2-1]\implies\boxed{S=\frac{\log2}{12}-\frac{1}{24}} $$

Method 2 $$ S=\frac{1}{24}\Big[-2[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots]+3[\frac{1}{3}+\frac{1}{5}+\dots]+[1+\frac{1}{3}+\frac{1}{5}+\dots]\Big]\\ =\frac{1}{24}\Big[-4[\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\dots]+4[1+\frac{1}{3}+\frac{1}{5}+\dots]-3\Big]\\ =\frac{1}{24}\Big[4[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots]+4.\frac{1}{2}-3\Big]=\frac{1}{24}\Big[4[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots]-1\Big]\\ =\frac{1}{24}[4\log2-1]\implies\boxed{S=\frac{\log2}{6}-\frac{1}{24}} $$ Why do I seem to get a different solution in Method 1 ?. And what is so different in methods 1 and 2 ?

Sooraj S
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    Rewriting the terms $T_n$ turns the sum from an absolutely convergent sum to a conditionally convergent sum whose terms cannot, in general, be rearranged. – Connor Harris Jun 27 '19 at 17:41
  • @ConnorHarris I do not have any deep understanding of wht u mentioned. I have edited OP to better explain my doubt, could u pls comment on that ? – Sooraj S Jun 28 '19 at 10:13
  • https://math.stackexchange.com/questions/13888/how-to-sum-frac11-cdot-2-cdot-3-cdot-4-frac43-cdot-4-cdot-5-cdot-6-f – V.G Mar 03 '21 at 11:55

3 Answers3

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Note that after the partial fraction decomposition we have that (one of your coefficients is wrong), $$\frac{1}{24}\left(\frac{-2}{n+1}+\frac{3}{2n+1}+\frac{\color{blue}{1}}{2n-1}\right)=\frac{1}{6}\left(\frac{-1}{2n+2}+\frac{1}{2n+1}\right)+\frac{1}{24}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right).$$ Hence, as $N\to\infty$, $$\begin{align} \sum_{n=1}^{N}T_n &=\frac{1}{6}\sum_{n=1}^{N}\left(\frac{-1}{2n+2}+\frac{1}{2n+1}\right) +\frac{1}{24}\sum_{n=1}^{N}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\\ &=\frac{1}{6}\sum_{n=3}^{2N+2}\frac{(-1)^{n+1}}{n} +\frac{1}{24}\left( 1-\frac{1}{2N+1}\right)\\ &\to\frac{1}{6}\left(\log(2)-1+\frac{1}{2}\right)+\frac{1}{24}=\frac{\log(2)}{6}-\frac{1}{24} \end{align}$$ where the last series is telescopic.

P.S. WA confirms the result. Note that in ss1729's first method $$\begin{align} &S_{2n}=\frac{1}{24}\Big[-2[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{2n+1}]+3[\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{4n+1}]+[1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{4n-1}]\Big]\\ &=\frac{1}{24}\Big[\color{red}{-2\big[\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n+1}\big]}+2\big[-\frac{1}{2}-\frac{1}{4}-\dots-\frac{1}{2n}\big]+\color{red}{2\big[\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n+1}+\frac{1}{2n+3}+\dots+\frac{1}{4n+1}\big]}+2\big[1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n+1}+\frac{1}{2n+3}+\dots+\frac{1}{4n+1}\big]-1-\frac{2}{4n+1}\Big] \end{align}$$ and the red part does NOT go to zero!!

Robert Z
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  • $$ S=\frac{1}{24}\Big[-2[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots]+3[\frac{1}{3}+\frac{1}{5}+\dots]+[1+\frac{1}{3}+\frac{1}{5}+\dots]\Big]\ =\frac{1}{24}\Big[\color{red}{-2\big[\frac{1}{3}+\frac{1}{5}+\dots\big]}+2\big[-\frac{1}{2}-\frac{1}{4}-\dots\big]+\color{red}{2\big[\frac{1}{3}+\frac{1}{5}+\dots\big]}+\big[\frac{1}{3}+\frac{1}{5}+\dots\big]+\big[1+\frac{1}{3}+\frac{1}{5}+\dots\big]\Big]\ =\frac{1}{24}\Big[2(-\frac{1}{2}-\frac{1}{4}-\dots)+2(1+\frac{1}{3}+\frac{1}{5}+\dots)-1\Big]\ =\frac{1}{24}[2\log2-1]\implies\boxed{S=\frac{\log2}{12}-\frac{1}{24}} $$ right? – Sooraj S Jun 27 '19 at 23:03
  • @ss1729 No, you are rearraging the terms of a non absolutely convergent series. In that way you change the final sum!! See my edit. – Robert Z Jun 28 '19 at 05:06
  • I have edited OP to include this doubt, could you please elaborate a bit more on why I seem to get a different solution in method 1 ? – Sooraj S Jun 28 '19 at 10:00
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    @ss1729 If you are going to rearrange the terms, then you should consider the finite partial sum and then take the limit (like in my edited answer). – Robert Z Jun 28 '19 at 11:17
  • alright. so is this the case with conditionally or non absolutely convergent series only ? – Sooraj S Jun 28 '19 at 16:39
  • Yes, if the series is unconditionally convergent you can rearrange the terms of the series. – Robert Z Jun 28 '19 at 16:42
  • $$ S=\frac{1}{24}\Big[-2[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n+1}]+3[\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n+1}]+[1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n-1}]\Big] =\frac{1}{24}\Big[-4[\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\dots\frac{1}{2n-1}+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}]+3[\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n-1}+\frac{1}{2n}+\frac{1}{2n+1}]+[1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n-1}]\Big] =\frac{1}{24}\Big[4[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2n-1}]-2+1-\frac{1}{2n}-\frac{1}{2n+1}-\frac{4}{2n+2}\Big] =\frac{1}{24}[4\log2-1] $$ – Sooraj S Jun 28 '19 at 17:15
  • isn't it the right way to approach ?. – Sooraj S Jun 28 '19 at 17:16
  • Yes, this approach is quite similar to my answer... – Robert Z Jun 28 '19 at 17:22
  • alright. thanx for the help. And could u please send some link whr I can read more about the problem that you mentioned about the conditionally or non absolutely convergent series ? – Sooraj S Jun 28 '19 at 17:27
  • @ss1729 See for example: https://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Brown135-138.pdf http://www.csun.edu/~hcmth017/riemann1/riemann1.pdf – Robert Z Jun 28 '19 at 20:44
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As Azif00 did, let us use partial fraction decomposition $$\frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}=\frac{1}{24(2n-1)}+\frac{1}{8(2n+1)}-\frac{1}{12(n+1)}$$ and then consider the partial sum $$S_p=\sum_{n=1}^p\frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}=\frac{1}{24}\sum_{n=1}^p\frac{1}{2n-1}+\frac{1}{8}\sum_{n=1}^p\frac{1}{2n+1}-\frac{1}{12}\sum_{n=1}^p\frac{1}{n+1}$$ Now, using harmonic numbers $$\sum_{n=1}^p\frac{1}{2n-1}=\frac{H_{p-\frac{1}{2}}}{2}+\log (2)$$ $$\sum_{n=1}^p\frac{1}{2n+1}=\frac{H_{p+\frac{1}{2}}}{2}-1+\log (2)$$ $$\sum_{n=1}^p\frac{1}{n+1}=H_{p+1}-1$$ Combining all the above $$S_p=\frac{H_{p-\frac{1}{2}}}{48}+\frac{H_{p+\frac{1}{2}}}{16}-\frac{H_{p+1}}{12}-\frac{ 1}{24}+\frac{\log (2)}{6}$$ Now, using the asymptotics of harmonic numbers $$S_p=\left(\frac{\log (2)}{6}-\frac{1}{24}\right)-\frac{1}{16 p}+\frac{1}{16 p^2}-\frac{13}{192 p^3}+O\left(\frac{1}{p^4}\right)$$ For a sanity check, $S_{10}=\frac{95219407}{1396755360}\approx 0.0681719$ while the truncated series gives $\frac{\log (2)}{6}-\frac{3031}{64000}\approx 0.0681652$.

Claude Leibovici
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Use partial fraction and we get $$\frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}=\frac{1}{24(2n-1)}+\frac{1}{8(2n+1)}-\frac{1}{6(2n+2)}$$ Thus \begin{align} \sum_{n=1}^{\infty}\frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}=&\sum_{n=1}^{\infty}\left(\frac{1}{24(2n-1)}+\frac{1}{8(2n+1)}-\frac{1}{6(2n+2)}\right) \\ =&\lim_{N\to \infty}\sum_{n=1}^N\left(\frac{1}{24(2n-1)}+\frac{1}{8(2n+1)}-\frac{1}{6(2n+2)}\right) \end{align} Use the telescoping series to find the partial sum and continue.

azif00
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