consider,$f(n)=\frac{1}{(2n-1) \cdot (2n-1+2)}$
where $n$ is a natural number
$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty\frac{1}{(2n-1) \cdot (2n+1)}$
Let, $\sum_{n=1}^{\infty} f(n) = S$
i.e. $S=\sum_{n=1}^\infty\frac{1}{(2n-1)\cdot(2n+1)}$ i.e. $S=\sum_{n=1}^\infty(\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1})$
i.e. $S=(\frac{1}{2})\left(\sum_{n=1}^\infty(\frac{1}{2n-1} - \frac{1}{2n+1})\right)$
i.e. $S=\lim_{n \to \infty}\left((\frac{1}{2})(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2})(\frac{1}{3} - \frac{1}{5}) + (\frac{1}{2})(\frac{1}{5} - \frac{1}{7}) + \cdots + (\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1})\right)$
i.e. $S=\lim_{n \to \infty}(\frac{1}{2})\left((\frac{1}{1} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + \cdots + (\frac{1}{2n-1} - \frac{1}{2n+1})\right)$
i.e. $S=\lim_{n \to \infty}(\frac{1}{2})\left(\frac{1}{1} - \frac{1}{2n+1}\right)$
i.e. $S=(\frac{1}{2})\left(\frac{1}{1} - \lim_{n \to \infty}(\frac{1}{2n+1})\right)$
i.e. $S=(\frac{1}{2})\left(\frac{1}{1} - 0\right)$
i.e. $S=(\frac{1}{2})\left(\frac{1}{1}\right)$
i.e. $S=\frac{1}{2}$