Answer: \(\frac{1}{(2r+1)(2r+3)}=\frac{1}{2}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)\).
Hence
\(\sum_{r=1}^{n}\frac{1}{(2r+1)(2r+3)}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2n+3}\right)=\frac{1}{6}-\frac{1}{2(2n+3)}\).
Therefore
\(\sum_{r=1}^{\infty}\frac{1}{(2r+1)(2r+3)}=\frac{1}{6}\).
Write the term as partial fractions:
\(\frac{1}{(2r+1)(2r+3)}=\frac{A}{2r+1}+\frac{B}{2r+3}\).
So
\(1=A(2r+3)+B(2r+1)\).
Comparing coefficients of \(r\): \(2A+2B=0\), so \(B=-A\). Comparing constants gives \(3A+B=1\), hence \(2A=1\) and \(A=\frac12\), \(B=-\frac12\). Therefore
\(\frac{1}{(2r+1)(2r+3)}=\frac{1}{2}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)\).
Now sum from \(r=1\) to \(n\):
\(\sum_{r=1}^{n}\frac{1}{(2r+1)(2r+3)}=\frac12\sum_{r=1}^{n}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)\).
Writing out the first few terms shows the cancellation:
\(\frac12\left(\left(\frac13-\frac15\right)+\left(\frac15-\frac17\right)+\cdots+\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\right)\).
Everything in the middle cancels, leaving
\(\sum_{r=1}^{n}\frac{1}{(2r+1)(2r+3)}=\frac12\left(\frac13-\frac{1}{2n+3}\right)=\frac16-\frac{1}{2(2n+3)}\).
As \(n\to\infty\), the final term tends to \(0\), so
\(\sum_{r=1}^{\infty}\frac{1}{(2r+1)(2r+3)}=\frac16\).
