Explanation

(a)
Here, $T_{n} = \sum_{n = 1}^{\infty}\frac{1}{(n + a)(n + 1 + a)}$
$= \sum_{n = 1}^{\infty}\left( \frac{1}{n + a} - \frac{1}{n + 1 + a} \right)$
$\therefore S_{n} = \Sigma T_{n} = \left( \frac{1}{1 + a} - \frac{1}{2 + a} \right) + \left( \frac{1}{2 + a} - \frac{1}{3 + a} \right)$
$+ ... + \left( \frac{1}{n + a} - \frac{1}{n + 1 + a} \right)$
$\Rightarrow S_{n} = \frac{1}{1 + a} - \frac{1}{n + 1 + a}$
$\Rightarrow \ \lim_{n \rightarrow \infty}{S_{n} = \frac{1}{1 + a}}$