Question

The sum of the series $1+\frac{1}{4\cdot 2!}+\frac{1}{16\cdot 4!}+\frac{1}{64\cdot 6!}+\cdots \infty$ is

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

Answer 1

The correct option is A $\frac{e+1}{2\sqrt{e}}$

We know that,
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$
$\Rightarrow e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\cdots$
$\Rightarrow \frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots$
On putting $x=\frac{1}{2}$, we get
$\frac{e^{1/2}+e^{-1/2}}{2}=1+{\left(\frac{1}{2}\right)}^2\frac{1}{2!}+{\left(\frac{1}{2}\right)}^4\frac{1}{4!}+\cdots$
$\Rightarrow \frac{e+1}{2\sqrt{e}}=1+{\left(\frac{1}{2}\right)}^2\frac{1}{2!}+{\left(\frac{1}{2}\right)}^4\frac{1}{4!}+\cdots$