The value of is: [2026]

Question

The value of $\sum_{k = 1}^{\infty} {(- 1)}^{k + 1} (\frac{k (k + 1)}{k!})$ is: [2026]

Smart Achievers · Accepted Answer

(4)

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

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

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

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

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

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

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

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

$= - [(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} . . . \infty) + 2 (- \frac{1}{0!} + \frac{1}{1!} - \frac{1}{2!} . . . \infty)]$

$= [\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} . . . \infty] = e^{- 1} = \frac{1}{e}$

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