Question

. The sum of the series $1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{64.6!}+\dots \dots \dots$ $\infty$. is

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

Answer 1

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

$\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\dots \dots$.
putting $x=1/2$ we get,
$\frac{e+1}{2\sqrt{e}}=1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{64.6!}+\dots \dots \dots \infty$