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$\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\ldots$ to $\infty$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{2 e}$
$n$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
$\begin{aligned}
&=\frac{1}{2}\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\frac{1}{6 !}-\frac{1}{7 !}+\ldots\right] \\
&=\frac{1}{2} e^{-1}=\frac{1}{2 e}
\end{aligned}$
$=\frac{1}{2}\left[\frac{(2 n+1)-1}{(2 n+1) !}\right]=\frac{1}{2}\left[\frac{1}{(2 n) !}-\frac{1}{(2 n+1) !}\right]$
$\therefore$ Sum of the series
$\begin{aligned}
&=\frac{1}{2}\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\frac{1}{6 !}-\frac{1}{7 !}+\ldots\right] \\
&=\frac{1}{2} e^{-1}=\frac{1}{2 e}
\end{aligned}$
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