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Question $$\sum_{n=1}^{\infty}\frac{n}{\left(2n-1\right)^{2}\left(2n+1\right)^{2}}$$

My Approach

$$\sum_{n=1}^{\infty}\frac{n}{\left(2n-1\right)^{2}\left(2n+1\right)^{2}}=\frac{1}{8}\left[\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^{2}}-\sum_{n=1}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\right]$$

$$S_{k}=\frac{1}{8}\left[\sum_{n=1}^{k}\frac{1}{\left(2n-1\right)^{2}}-\sum_{n=1}^{k}\frac{1}{\left(2n+1\right)^{2}}\right]$$

now i don't know it is right to replace n by $\frac{n}{2}$

But doing that,

$$S_{k}=\frac{1}{8}\left[\sum_{n=1}^{2k-1}\frac{1}{\left(n\right)^{2}}-\sum_{n=3}^{2k+1}\frac{1}{\left(n\right)^{2}}\right]=\frac{1}{8}\left[1+\frac{1}{4}-\frac{1}{\left(2k\right)^{2}}-\frac{1}{\left(2k+1\right)^{2}}\right]$$

$$ \lim_{k\rightarrow\infty}S_{k}=\frac{5}{32}$$

Book's answer

Similarly, $a_n=\frac{1}{8} \left( \frac{1}{(2n-1)^2}-\frac{1}{(2n+1)^2} \right)$. It then follows that $S_n= \left( 1-\frac{1}{(2n+1)^2} \right)$ and $S= \lim_{n \to \infty} S_n=\frac{1}{8}$.

Reference Request

Definitely i have mistaken somewhere. I don't know many things about these type of problems. My book only contains problems and very brief solutions and no theory. Any book suggestion will be very useful.

Yuriy S
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Mohan Sharma
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1 Answers1

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Use $$\frac{n}{(2n-1)^2(2n+1)^2}=\frac{1}{8}\left(\frac{1}{(2n-1)^2}-\frac{1}{(2n+1)^2}\right)$$ and the telescopic sum.

Thus, we have $$S_n=\frac{1}{8}\left(1-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+...+\frac{1}{(2n-1)^2}-\frac{1}{(2n+1)^2}\right)=$$ $$=\frac{1}{8}\left(1-\frac{1}{(2n+1)^2}\right)\rightarrow\frac{1}{8}.$$

Michael Rozenberg
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