Question

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

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

Answer 1

Answer: (A)

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