13!+25!+37!+……∞=

13+25+37+……∞ =∑1∞ n(2n+1) =12∑1∞ (2n+1)-12n+1 =12∑1∞ 12n-12n+1 =1212-13+14-15+.... =12e-1

13!+25!+37!+……∞=

13+25+37+……∞ =∑1∞ n(2n+1) =12∑1∞ (2n+1)-12n+1 =12∑1∞ 12n-12n+1 =1212-13+14-15+.... =12e-1

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$\frac{1}{3!} + \frac{2}{5!} + \frac{3}{7!} + \dots \dots \infty =$

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$\frac{1}{2 e}$

$\frac{e}{2}$

$\frac{3}{2 e}$

$\frac{1}{e}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} \frac{1}{3} + \frac{2}{5} + \frac{3}{7} + \dots \dots \infty \\ = & \sum_{1}^{\infty} \frac{n}{(2 n + 1)} \\ = & \frac{1}{2} (\sum_{1}^{\infty} \frac{(2 n + 1) - 1}{2 n + 1}) \\ = & \frac{1}{2} (\sum_{1}^{\infty} \frac{1}{2 n} - \frac{1}{2 n + 1}) \\ = & \frac{1}{2} (\frac{1}{2}) - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + . . . . \\ = & \frac{1}{2} (e^{- 1}) \end{array}$