Answer
$$\frac{1}{2}$$
Work Step by Step
Given \begin{align*}
\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\cdots&=\sum_{n=1}^{n=\infty} \frac{1}{(2 n-1)(2 n+1)}\\
&=\sum_{n=1}^{n=\infty} \frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)\end{align*}
Since
\begin{aligned} S_{N} &=\sum_{n=1}^{N} \frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right) \\ &=\frac{1}{2} \sum_{n=1}^{N}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right) \\ &=\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\ldots+\left(\frac{1}{2 N-x}-\frac{1}{2 N+1}\right)\right] \\ &=\frac{1}{2}\left(1-\frac{1}{2 N+1}\right) \end{aligned}
Then
$$ S=\lim_{N\to\infty} S_N=\frac{1}{2}$$
