Find $\frac{1}{1.2.3.4}+\frac{4}{3.4.5.6}+\frac{9}{5.6.7.8}+\frac{16}{7.8.9.10}+\dots$

Question

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 ?

Answer 1

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!!

Answer 2

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.

Answer 3

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 $\frac{\log (2)}{6}-\frac{3031}{64000}\approx 0.0681652$.