sum_{k=1}^{infty} frac{1}{k!}left(sum_{n=1}^k | vedclass

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(D) The inner sum is a geometric series: $\sum_{n=1}^k 2^{n-1} = \frac{1(2^k - 1)}{2 - 1} = 2^k - 1$. Substituting this into the main expression: $\su

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$e^2-e$

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(D) The inner sum is a geometric series: $\sum_{n=1}^k 2^{n-1} = \frac{1(2^k - 1)}{2 - 1} = 2^k - 1$. Substituting this into the main expression: $\sum_{k=1}^{\infty} \frac{2^k - 1}{k!} = \sum_{k=1}^{\infty} \frac{2^k}{k!} - \sum_{k=1}^{\infty} \frac{1}{k!}$. We know the Taylor series expansion $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$,so $\sum_{k=1}^{\infty} \frac{x^k}{k!} = e^x - 1$. For the first part,$x = 2$: $\sum_{k=1}^{\infty} \frac{2^k}{k!} = e^2 - 1$. For the second part,$x = 1$: $\sum_{k=1}^{\infty} \frac{1^k}{k!} = e^1 - 1 = e - 1$. Subtracting these: $(e^2 - 1) - (e - 1) = e^2 - 1 - e + 1 = e^2 - e$.

$\sum_{k=1}^{\infty} \frac{1}{k!}\left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

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