Question

The sum of the infinite series
$1+\frac{1}{2!}+\frac{1\cdot3}{4!}+\frac{1\cdot3\cdot5}{6!}+\ldots$ is

• A. $e$
• B. $e^{2}$
• C. $\sqrt{e}$
• D. $\frac{1}{e}$

Answer 1

Answer: (C)

Let $S=1+\frac{1}{2!}+\frac{1\cdot3}{4!}+\frac{1\cdot3\cdot5}{6!}+ \ldots\infty$

$\therefore T_{n}=\frac{1\cdot3\cdot5\cdot\ldots\cdot\left(2n-1\right)}{\left(2n\right)!}\times \frac{2\cdot4\cdot\ldots\cdot2n}{2\cdot4\cdot\ldots\cdot2n}$

$=\frac{\left(2n\right)!}{\left(2n\right)!2^{n} \left(n\right)!}=\frac{1}{2^{n} \left(n\right)!}$

$\therefore S=1+\sum T_{n}=1+\frac{1}{2\left(1\right)!}+\frac{1}{2^{2}\left(2\right)!}+\infty$

$=e^{1/ 2}=\sqrt{e}$