Question

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

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

Answer 1

The correct option is C

$e^{\frac{1}{2}}$

Explanation for the correct option:

Given, $1+\frac{1}{2!}+\frac{1·3}{4!}+\frac{1·3·5}{6!}+\dots$

Now, $n^{\mathrm{th}}$ term of this series$=\frac{1·3·...·(2n-1)}{2n!}$ (excluding first term)

$\begin{array}{rcl}& \Rightarrow & T_n=\frac{1·3·...·(2n-1)}{2n!}\times \frac{(2·4·...·2n)}{(2·4·...·2n)}\\ & \Rightarrow & T_n=\frac{2n!}{2n!\times 2^{n}n!}\\ & =& \frac{1}{2^{n}n!}\end{array}$

So,

$\Rightarrow S_n=1+\frac{1}{2\times 1!}+\frac{1}{2^2\times 2!}+\frac{1}{2^3\times 3!}+...$

We know that,

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

When $x=\frac{1}{2}$,

$\begin{array}{rcl}{e}^{\frac{1}{2}}& =& 1+\frac{{\displaystyle \frac{1}{2}}}{1!}+\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2}{2!}+\frac{{\left({\displaystyle \frac{1}{2}}\right)}^3}{3!}+...\\ & =& 1+\frac{1}{2\times 1!}+\frac{1}{2^2\times 2!}+\frac{1}{2^3\times 3!}+...\\ & =& S_n\end{array}$

$\therefore S_n=e^{\frac{1}{2}}$

Hence, option C is correct.