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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 :

Similar Questions

$3 + \frac{5}{1!} + \frac{7}{2!} + \frac{9}{3!} + \dots \infty = $

The coefficient of $x^{6}$ in the expansion of $e^{2x}$ is:

The sum of the series $\frac{1^2}{1 \cdot 2!} + \frac{1^2 + 2^2}{2 \cdot 3!} + \frac{1^2 + 2^2 + 3^2}{3 \cdot 4!} + \dots + \frac{1^2 + 2^2 + \dots + n^2}{n(n + 1)!} + \dots \infty$ is equal to:

$1+\frac{1+2}{2 !}+\frac{1+2+2^2}{3 !}+\ldots$ is equal to

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8 \times 2!} + \frac{1}{16 \times 3!} + \frac{1}{32 \times 4!} + \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module