Question

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

• A. \( \dfrac{1}{e} \)
• B. \( \dfrac{2}{e} \)
• C. \( \sqrt{e} \)
• D. \( \dfrac{e}{2} \)

Hint:

Always try to rewrite factorial expressions to resemble the series of \( e^x \), \( xe^x \), or their derivatives.

Answer 1

The Correct Option is B

Concept: Series involving factorials are often simplified by rewriting terms to match known expansions of \( e^x \).
Step 1: Rewrite the general term \[ \frac{k(k+1)}{k!} = \frac{k^2 + k}{k!} = \frac{k}{(k-1)!} + \frac{1}{(k-1)!} \] Thus, \[ \sum_{k=1}^{\infty} (-1)^{k+1}\frac{k(k+1)}{k!} = \sum_{k=1}^{\infty} (-1)^{k+1}\left(\frac{k}{(k-1)!} + \frac{1}{(k-1)!}\right) \]
Step 2: Split the series \[ = \sum_{k=1}^{\infty} (-1)^{k+1}\frac{k}{(k-1)!} + \sum_{k=1}^{\infty} (-1)^{k+1}\frac{1}{(k-1)!} \] Let \( n = k-1 \): \[ = \sum_{n=0}^{\infty} (-1)^n \frac{n+1}{n!} + \sum_{n=0}^{\infty} (-1)^n \frac{1}{n!} \]
Step 3: Use known expansions \[ \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = e^{-1} \] \[ \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)}{n!} = 0 \] Thus, \[ \text{Required sum} = 2e^{-1} = \frac{2}{e} \]