What is the limiting sum of $\frac{1}{1\left(3\right)}+\frac{1}{3\left(5\right)}+\frac{1}{5\left(7\right)}+···+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ I know $a$ (first term) is $\frac{1}{3}$, but I'm not sure what the common ratio would be from this. I assume it is the $\frac{1}{(2n-1)(2n+1)}$, but this is not a number. Any hints?

$$S=\frac12\big(\frac{1}{2-1}-\frac{1}{2+1}+\frac{1}{4-1}-\frac{1}{4+1}+ \frac{1}{6-1}-\frac{1}{6+1}+\cdot\cdot\cdot+\frac{1}{2n-1}-\frac{1}{2n+1}\big) $$
Cancelling similar terms we finally get:
$$S= \frac12\big(1-\frac{1}{2n+1}\big)=\frac12\times\frac{2n}{2n+1}$$
$$\lim _{n→∞ } S=\frac{1}{2}$$
