Q84 / 180 · COMEDK 2017 · Mathematics

$\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\ldots$ to $\infty$ is equal to

  • A
    $e$
  • B
    $2 e$
  • Correct answer
    C
    $\frac{1}{2 e}$
  • D
    $\frac{e}{2}$

Solution

$n$th term $=\frac{n}{(2 n+1) !}$ $=\frac{1}{2}\left[\frac{(2 n+1)-1}{(2 n+1) !}\right]=\frac{1}{2}\left[\frac{1}{(2 n) !}-\frac{1}{(2 n+1) !}\right]$ $\therefore$ Sum of the series $\begin{aligned} &=\frac{1}{2}\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\frac{1}{6 !}-\frac{1}{7 !}+\ldots\right] \\ &=\frac{1}{2} e^{-1}=\frac{1}{2 e} \end{aligned}$

Hence, the correct answer is option: 'C'.

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