Question

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

• A. $\frac{e-1}{\sqrt{e}}$
• B. $\frac{e+1}{\sqrt{e}}$
• C. $\frac{e-1}{2\sqrt{e}}$
• D. None of these

Answer 1

The correct option is D None of these

$1+\frac{1}{4(2!)}+\frac{1}{16(4!)}+\frac{1}{64(6!)}..........\infty$

$=1+\frac{1}{2^2(2!)}+\frac{1}{2^4(4!)}+\frac{1}{2^6(6!)}........\infty$

$=\frac{1}{2}\left[2(1+\frac{1}{2^2(2!)}+\frac{1}{2^4(4!)}+\frac{1}{2^6(6!)}.......\infty )\right]$

$=\frac{1}{2}\left[2(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}......\infty )\right]$

$=\frac{1}{2}[e^x+e^{-x}]$, where $x=\frac{1}{2}$

$=\frac{1}{2}[e^{1/2}+e^{-1/2}]$

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