Answer: \(\displaystyle \sum_{r=1}^{n}\frac{1}{(2r)^2-1}=\frac12\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}\).
Therefore \(\displaystyle \sum_{r=1}^{\infty}\frac{1}{(2r)^2-1}=\frac12\).
First factor the denominator:
\((2r)^2-1=(2r-1)(2r+1).\)
Now use partial fractions:
\(\frac{1}{(2r-1)(2r+1)}=\frac12\left(\frac{1}{2r-1}-\frac{1}{2r+1}\right).\)
Hence
\(\sum_{r=1}^{n}\frac{1}{(2r)^2-1}=\frac12\sum_{r=1}^{n}\left(\frac{1}{2r-1}-\frac{1}{2r+1}\right).\)
Writing out the terms gives
\(\frac12\left[\left(1-\frac13\right)+\left(\frac13-\frac15\right)+\cdots+\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\right].\)
All the middle terms cancel, so
\(\sum_{r=1}^{n}\frac{1}{(2r)^2-1}=\frac12\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}.\)
Letting \(n\to\infty\),
\(\sum_{r=1}^{\infty}\frac{1}{(2r)^2-1}=\frac12.\)
