Question

The value of the series $\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+.....$ to $\infty$ is equal to

• A. $\frac{1}{2}e$
• B. $\frac{1}{2e}$
• C. $\frac{3}{2e}$
• D. $\frac{4}{5e}$

Answer 1

The correct option is B $\frac{1}{2e}$

$\frac{0}{1!}+\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+\frac{4}{9!}+...=S$

$S={\sum }_{k=0}^{\infty }\frac{k}{(2k+1)!}$

$=\frac{1}{2}{\sum }_{k=0}^{\infty }\frac{2k}{(2k+1)!}$

$=\frac{1}{2}{\sum }_{k=0}^{\infty }\frac{2k+1-1}{(2k+1)!}$

$=\frac{1}{2}[{\sum }_{k=0}^{\infty }\frac{2k+1}{(2k+1)!}-{\sum }_{k=0}^{\infty }\frac{1}{(2k+1)!}]$

$=\frac{1}{2}[{\sum }_{k=0}^{\infty }\frac{1}{\left(2k\right)!}-{\sum }_{k=0}^{\infty }\frac{1}{(2k+1)!}]$

$=\frac{1}{2}{\sum }_{k=0}^{\infty }\frac{1}{\left(2k\right)!}-\frac{1}{2}{\sum }_{k=0}^{\infty }\frac{1}{(2k+1)!}$

Now,$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$

$e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...$

$\Rightarrow e+e^{-1}=2[1+\frac{1}{2!}+\frac{1}{4!}+...]=2{\sum }_{k=1}^{\infty }\frac{1}{\left(2k\right)!}$

$\Rightarrow {\sum }_{k=1}^{\infty }\frac{1}{\left(2k\right)!}=\frac{1}{2}(e+e^{-1})$

$e-e^{-1}=2[1+\frac{1}{3!}+\frac{1}{5!}+...]=2{\sum }_{k=1}^{\infty }\frac{1}{(2k+1)!}$

$\Rightarrow {\sum }_{k=1}^{\infty }\frac{1}{(2k+1)!}=\frac{1}{2}(e-e^{-1})$

$\therefore S=\frac{1}{2}\left(\frac{e+e^{-1}}{2}\right)-\frac{1}{2}\left(\frac{e-e^{-1}}{2}\right)$

$=\frac{1}{4}(e+e^{-1}-e+e^{-1})$

$=\frac{1}{4}\left(2e^{-1}\right)$

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