Question

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

Answer: (D)

We know that,
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....$
and $e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-....$
$\Rightarrow$ $\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+....$
Put $x=\frac{1}{2}$
$\frac{e^{1/2}+e^{-1/2}}{2}=1+{\left(\frac{1}{2}\right)}^2.\frac{1}{2!}+{\left(\frac{1}{2}\right)}^4\frac{1}{4!}+....$
$\Rightarrow$ $\frac{e+1}{2\sqrt{e}}=1{\left(\frac{1}{2}\right)}^2.\frac{1}{2!}+{\left(\frac{1}{2}\right)}^4\frac{1}{4!}+...$