Question

The sum of the infinite series $1+\frac{1}{2!}+\frac{1\cdot 3}{4!}+\frac{1\cdot 3\cdot 5}{6!}+\dots$ is

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

Answer 1

Answer: (C)

$\sqrt{e}$

Let $S=1+\frac{1}{2!}+\frac{1\cdot 3}{4!}+\frac{1\cdot 3\cdot 5}{6!}+\dots \infty$
$\therefore T_n=\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)}{\left(2n\right)!}\times \frac{2\cdot 4\cdot \cdots \cdot 2n}{2\cdot 4\cdot \cdots \cdot 2n}$
$=\frac{\left(2n\right)}{\left(2n\right)!2^n\left(n\right)!}=\frac{1}{2^n\left(n\right)!}$
$\therefore S=1+\sum T_n=1+\frac{1}{2\left(1\right)!}+\frac{1}{2^2\left(2\right)!}+\dots \infty$
$=e^{1/2}=\sqrt{e}$