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According to Wolfram Alpha, $$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)(2n-1)}=\ln(4)-1$$ However, I am not sure how to evaluate this series.

Attempt $$\frac{1}{n(2n+1)(2n-1)}=\frac{A}{n}+\frac{B}{2n+1}+\frac{C}{2n-1}$$ $$1=A(2n+1)(2n-1)+Bn(2n-1)+Cn(2n+1)$$

Then, I got $$ \sum_{n=1}^{\infty}\frac{1}{n(2n+1)(2n-1)} = \sum_{n=1}^{\infty}\frac{-1}{n}+\frac{1}{2n+1}+\frac{1}{2n-1}\\ $$ I tried to view this as a telescoping series, but it did not turn out good. Can I have a hint?

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$$ \sum_{n\geq 1} \left( \frac{1}{2n-1} - \frac{1}{2n}\right) - \sum_{n\geq 1} \left( \frac{1}{2n} - \frac{1}{2n+1}\right) = \ln 2 - (1-\ln 2) $$

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