Explanation

(d)
$\because T_{n} = \frac{5 + (n - 1)4}{\left\lbrack 3 + (n - 1)4 \right\rbrack^{2}\left\lbrack 7 + (n - 1)4 \right\rbrack^{2}\ }$
$= \frac{1}{8}\ \left\{ \frac{1}{(4n - 1)^{2}} - \frac{1}{(4n + 3)^{2}} \right\}$
$\therefore S_{n} = T_{1} + T_{2} + ... + T_{n}$
$= \frac{1}{8}\ \left\{ \frac{1}{3^{2}} - \frac{1}{7^{2}} + \frac{1}{7^{2}} = \frac{1}{11^{2}} + \frac{1}{11^{2}} - ... + \frac{1}{(4n - 1)^{2}} - \frac{1}{(4n + 3)^{2}} \right\}$
$= \frac{1}{8}\left\{ \frac{1}{3^{2}} - \frac{1}{(4n + 3)^{2}} \right\}$
$\Rightarrow S_{\infty} = \frac{1}{8}\left\{ \frac{1}{9} - 0 \right\} = \frac{1}{72}$