11!+1+32!+1+3+53!+……∞=

11+1+32+1+3+53+…… =∑1∞ 1+3+5+...+(2n-1)n =∑1∞ n2n=∑1∞ nn-1 =∑1∞ (n-1)+1n-1 =∑1∞.1n-2+1n-1 =e+e=2e

11!+1+32!+1+3+53!+……∞=

11+1+32+1+3+53+…… =∑1∞ 1+3+5+...+(2n-1)n =∑1∞ n2n=∑1∞ nn-1 =∑1∞ (n-1)+1n-1 =∑1∞.1n-2+1n-1 =e+e=2e

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

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$\sqrt{e}$

$\frac{e}{2}$

answer is B.

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Detailed Solution

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