Question

The sum of the series $1+\frac{1}{4\cdot2!}+\frac{1}{16\cdot 4!}+\frac{1}{64\cdot 6!} + \ldots \infty$ 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)

$\frac{e+1}{2\sqrt{e}}$

Answer 2

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