jmc

The $n$th term of the series is given by $$\frac{4n+1}{(4n-1{)}^2(4n+3{)}^2}.$$Note that $$\begin{array}{rl}(4n+3{)}^2-(4n-1{)}^2& =[(4n+3)+(4n-1)][(4n+3)-(4n-1)]\\ & =(8n+2)(4)=8(4n+1),\end{array}$$so we can write $$\begin{array}{rl}\frac{4n+1}{(4n-1{)}^2(4n+3{)}^2}& =\frac{1}{8}\left[\frac{(4n+3{)}^2-(4n-1{)}^2}{(4n-1{)}^2(4n+3{)}^2}\right]\\ & =\frac{1}{8}\left(\frac{1}{(4n-1{)}^2}-\frac{1}{(4n+3{)}^2}\right).\end{array}$$Thus, $$\begin{array}{rl}\frac{5}{3^2\cdot 7^2}+\frac{9}{7^2\cdot 11^2}+\frac{13}{11^2\cdot 15^2}+\cdots & =\frac{1}{8}\left(\frac{1}{3^2}-\frac{1}{7^2}\right)+\frac{1}{8}\left(\frac{1}{7^2}-\frac{1}{11^2}\right)+\frac{1}{8}\left(\frac{1}{11^2}-\frac{1}{15^2}\right)+\cdots \\ & =\frac{1}{8}\cdot \frac{1}{3^2}=\boxed{\frac{1}{72}}.\end{array}$$