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 ?
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