Answer
$\frac{1}{6}$, Converges
Work Step by Step
Begin by splitting into partial fractions
$\Sigma_{n=1}^{\infty} \frac{1}{(2n+1)(2n+3)}$
$\frac{1}{(2n+1)(2n+3)} = \frac{A}{2n+1}+ \frac{B}{2n+3}$
$1=A(2n+3) +B(2n+1)$
$A=-B$
$3A+B=1$
$B=-\frac{1}{2}$, $A=\frac{1}{2}$
Resubstitute
$\Sigma_{n=1}^{\infty} \frac{1}{2(2n+1)}-\frac{1}{2(2n+3)}$
Write out the first few terms
$(\frac{1}{6}-\frac{1}{10}) + (\frac{1}{10}-\frac{1}{14})+(\frac{1}{14}-\frac{1}{18})$
Rewrite the terms of the sum as a limit
$\lim\limits_{n \to \infty} \frac{1}{6}-\frac{1}{2(2n+3)}$
Evaluate the limit
$\frac{1}{6}$, Converges