Explanation

(b)
Let $S = \frac{1}{1.3} + \frac{1}{2.5} + \frac{1}{3.7} + \frac{1}{4.9} + ...$
$\therefore T_{n} = \frac{1}{n(2n + 1)} = \frac{1}{n} - \frac{2}{(2n + 1)}$
$\Rightarrow S = \sum_{n = 1}^{\infty}T_{n} = \frac{1}{1} - \frac{2}{3} + \frac{1}{2} - \frac{2}{5} + \frac{1}{3} - \frac{2}{7} + \frac{1}{4} - \frac{2}{9} + \frac{1}{5} - ...$
$= 1 - \left( - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ... \right)$
$= 1 - \left( - 1 + \log_{e}2 \right)$
$= 2 - \log_{e}2$