Explanation

(a)
Let $S = \frac{1}{1.2.3} + \frac{1}{3.4.5} + \frac{1}{5.6.7} + ...\infty$
$\therefore\ T_{n} = \frac{1}{(2n - 1)(2n)(2n + 1)}$
$= \frac{1}{2(2n - 1)} - \frac{1}{2n} + \frac{1}{2(2n + 1)}$
$= \frac{1}{2}\left\lbrack \frac{1}{2n - 1} - \frac{1}{2n} \right\rbrack - \frac{1}{2}\left\lbrack \frac{1}{2n} - \frac{1}{2n + 1} \right\rbrack$
On putting $n = 1,2,3,\ldots$
$T_{1} = \frac{1}{2}\left\lbrack \frac{1}{1} - \frac{1}{2} \right\rbrack - \frac{1}{2}\left\lbrack \frac{1}{2} - \frac{1}{3} \right\rbrack$
$T_{2} = \frac{1}{2}\left\lbrack \frac{1}{3} - \frac{1}{4} \right\rbrack - \frac{1}{2}\left\lbrack \frac{1}{4} - \frac{1}{5} \right\rbrack$
$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$
$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$
$\therefore\ S = T_{1} + T_{2} + T_{3} + ... + T_{n} + ...$
$= \frac{1}{2}\left\lbrack 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - ... \right\rbrack - \frac{1}{2}\left\lbrack \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \frac{1}{6} - \frac{1}{7} + ... \right\rbrack$
$= \frac{1}{2}\log_{e}{(1 + 1)} + \frac{1}{2}\left\lbrack - 1 + \left\{ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \right\} \right\rbrack$
$= \frac{1}{2}\log_{e}2 - \frac{1}{2} + \frac{1}{2}\log_{e}{(1 + 1)}$
$= \log_{e}2 - \frac{1}{2}$