Explanation

(b)
We have, $\sum_{n = 1}^{\infty}{\frac{2n}{(2n + 1)!} = \sum_{n = 1}^{\infty}\frac{2n + 1 - 1}{(2n + 1)!}}$
$= \sum_{n = 1}^{\infty}\left( \frac{1}{(2n)!} - \frac{1}{(2n + 1)!} \right)$
$\sum_{n = 1}^{\infty}{\frac{1}{(2n)!} - \sum_{n = 1}^{\infty}\frac{1}{(2n + 1)!}}$
$= \left\lbrack \frac{e + e^{- 1}}{2} - 1 \right\rbrack - \left\lbrack \frac{e - e^{- 1}}{2} - 1 \right\rbrack = e^{- 1}$