The value of sum_{k=1}^{infty}(-1)^{k+1}(frac | vedclass

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Question

(B) We are given the series $S = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{k(k+1)}{k!}$.Note that $\frac{k(k+1)}{k!} = \frac{k(k-1+2)}{k!} = \frac{k(k-1)}{k!

Vedclass · Accepted Answer

$1/e$

Vedclass · Answer

(B) We are given the series $S = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{k(k+1)}{k!}$.Note that $\frac{k(k+1)}{k!} = \frac{k(k-1+2)}{k!} = \frac{k(k-1)}{k!} + \frac{2k}{k!} = \frac{1}{(k-2)!} + \frac{2}{(k-1)!}$ for $k \ge 2$.For $k=1$, the term is $(-1)^{1+1}\frac{1(2)}{1!} = 2$.Expanding the sum: $S = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k(k-1)}{k!} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{2k}{k!}$.$S = \sum_{k=2}^{\infty} \frac{(-1)^{k+1}}{(k-2)!} + 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(k-1)!}$.Let $j = k-2$ in the first sum and $m = k-1$ in the second sum:$S = -\sum_{j=0}^{\infty} \frac{(-1)^{j}}{j!} + 2 \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!}$.Since $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = e^{-1} = \frac{1}{e}$, we have:$S = -(\frac{1}{e}) + 2(\frac{1}{e}) = \frac{1}{e}$.

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

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