∑k=1∞1k!∑n=1k2n-1=

∑k=1∞ 1k ∑n=1k2n-1 =∑k=1∞ 1k1+2+22+....+2k-1 =∑k=1∞ 1k 2k-1(2-1) =∑k=1∞ 2k-1k =∑k=1∞ 2kk-∑k=1∞ 1k =e2-1-(e-1) =e2-e

∑k=1∞1k!∑n=1k2n-1=

∑k=1∞ 1k ∑n=1k2n-1 =∑k=1∞ 1k1+2+22+....+2k-1 =∑k=1∞ 1k 2k-1(2-1) =∑k=1∞ 2k-1k =∑k=1∞ 2kk-∑k=1∞ 1k =e2-1-(e-1) =e2-e

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$\sum_{k = 1}^{\infty} \frac{1}{k!} (\sum_{n = 1}^{k} 2^{n - 1}) =$

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$e^{2} + e$

$e^{2}$

$e^{2} - e$

answer is D.

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Detailed Solution

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