• # question_answer The sum of the series $1+\frac{3}{2\,!}+\frac{7}{3\,!}+\frac{15}{4\,!}+.....\text{to}\,\infty$ is  [Kerala (Engg.) 2005] A) $e(e+1)$ B) $e\,(1-e)$ C) $3e-1$ D) $3e$ E) (e) $e\,(e-1)$

(e) $1+\frac{3}{2\,!}+\frac{7}{3\,!}+\frac{15}{4\,!}+....$ $=\,(1-1)+\left( \frac{2}{1!}-\frac{1}{1!} \right)+\left( \frac{{{2}^{2}}}{2\,!}-\frac{{{1}^{2}}}{2\,!} \right)+\left( \frac{{{2}^{3}}}{3\,!}-\frac{{{1}^{3}}}{3\,!} \right)+...$ $=\left( 1+\frac{2}{1!}+\frac{{{2}^{2}}}{2!}+\frac{{{2}^{3}}}{3!}+.... \right)$$-\left( 1+\frac{1}{1!}+\frac{{{1}^{2}}}{2!}+\frac{{{1}^{3}}}{3!}+... \right)$ $={{e}^{2}}-e$ = $e(e-1)$.