Answer
See proof
Work Step by Step
\begin{aligned}
\textbf{Step 1}\
&\text{Given the equation}\\
&\frac{1}{1\cdot 3} + \frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \cdots = \frac{1}{2}\\
&\text{and we're asked to show that.}\\
\textbf{Step 2}\\
&\text{Let } s_n \text{ denote the sum of the initial terms of the series, up to and including the term with index } n.\\
&\text{As } n \text{ increases, the partial sum } s_n \text{ includes more and more terms of the series, thus}\\
&\quad \sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)} = \frac{1}{1\cdot 3} + \frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \cdots + \frac{1}{(2n-1)(2n+1)}\\
\\
\textbf{Step 3}\\
&\text{Let's write } s_n \text{ in the closed form by using the method of partial fractions.}\\
&\frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1}\\
&\frac{1}{(2k-1)(2k+1)} = \frac{(2k+1)A + (2k-1)B}{(2k-1)(2k+1)}\\
&\begin{cases}
A+B = 0\\
A - B = 1
\end{cases}\\
&\Rightarrow A = \frac{1}{2}, B = -\frac{1}{2}\\
\\
\textbf{Step 4}\\
&\text{Thus,}\\
& s_n = \frac{1}{2} \sum_{k=1}^{n} \frac{1}{2k-1} - \frac{1}{2} \sum_{k=1}^{n} \frac{1}{2k+1}\\
&\Rightarrow s_n = \frac{1}{2} \left(1 - \frac{1}{2n+1}\right)\\
\\
\textbf{Step 5}\\
&\text{Therefore,}\\
&\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} = \lim_{n \to \infty} s_n = \frac{1}{2} \left(1 - \lim_{n \to \infty} \frac{1}{2n+1}\right)\\
&= \frac{1}{2} \cdot 1 = \frac{1}{2}\\
\end{aligned}
