Explanation

(a)
Here, $T_{n} = \frac{n}{(2n + 1)!} = \frac{1}{2}\left\lbrack \frac{2n + 1 - 1}{(2n + 1)!} \right\rbrack$
$= \frac{1}{2}\left\lbrack \frac{1}{(2n)!} - \frac{1}{(2n + 1)!} \right\rbrack$
$\therefore T_{1} = \frac{1}{2}\left( \frac{1}{2!} - \frac{1}{3!} \right)$
$T_{2} = \frac{1}{2}\left( \frac{1}{4!} - \frac{1}{5!} \right)$
$\cdots\ \ \ \cdots\ \ \ \ \cdots$
$\cdots\ \ \ \cdots\ \ \ \cdots$
$\cdots\ \ \ \cdots\ \ \ \cdots$
$\therefore S = \sum_{n = 1}^{\infty}{T_{n} = \frac{1}{2}\left\lbrack \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + ...\infty + 1 - 1 \right\rbrack = \frac{e^{- 1}}{2}}$