Question

The sum of the series $1+\frac{1}{4\times 2!}+\frac{1}{16\times 4!}+\frac{1}{64\times 6!}+....\infty$ 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

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

given series is $1+\frac{1}{4\times 2!}+\frac{1}{16\times 4!}+\frac{1}{64\times 6!}+....\infty$

We know that $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....\infty$

Let the series be denoted by $S$

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

$=(1+\frac{1}{1!}(\frac{1}{2})+\frac{1}{2!}{\left(\frac{1}{2}\right)}^2+\frac{1}{3!}{\left(\frac{1}{2}\right)}^2+....\infty )-(\frac{1}{1!}{\left(\frac{1}{2}\right)}^2+\frac{1}{3!}{\left(\frac{1}{2}\right)}^3+....\infty )$

$2S=(1+\frac{1}{1!}(\frac{1}{2})+\frac{1}{2!}{\left(\frac{1}{2}\right)}^2+....\infty )+(1-\frac{1}{1!}(\frac{1}{2})+\frac{1}{2!}{\left(\frac{1}{2}\right)}^2-\frac{1}{3!}{\left(\frac{1}{2}\right)}^3+....\infty )$

By $e^x$ expansion $e^{1/2}=1+\frac{1}{1!}\left(\frac{1}{2}\right)+\frac{1}{2!}{\left(\frac{1}{2}\right)}^2+....\infty )$

$e^{-1/2}=1-\frac{1}{1!}\left(\frac{1}{2}\right)+\frac{1}{2!}{\left(\frac{1}{2}\right)}^2+...\infty$

$2S=e^{1/2}+e^{-1/2}$

$S=\cfrac { \sqrt { e } +\cfrac { 1 }{ \sqrt { e } } }{ 2 } =\cfrac { e+1 }{ 2\sqrt { e } }