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

Step 1: Understanding the Concept:
This is an infinite series problem involving factorials and alternating signs, which points toward the expansion of \(e^x\).
Step 2: Key Formula or Approach:
Use the expansion \(e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \dots\).
Step 3: Detailed Explanation:
General term \(T_k = (-1)^{k+1} \frac{k(k+1)}{k!} = (-1)^{k+1} \frac{k+1}{(k-1)!}\).
Write \(k+1 = (k-1) + 2\):
\(T_k = (-1)^{k+1} \left[ \frac{k-1}{(k-1)!} + \frac{2}{(k-1)!} \right] = \frac{(-1)^{k+1}}{(k-2)!} + \frac{2(-1)^{k+1}}{(k-1)!}\).
Summing from \(k=1\) to \(\infty\):
For the first part \(\sum_{k=1}^\infty \frac{(-1)^{k+1}}{(k-2)!}\): The terms exist for \(k \geq 2\).
Let \(n = k-2\). As \(k \to 2, n \to 0\).
\(\sum_{n=0}^\infty \frac{(-1)^{n+3}}{n!} = -\sum_{n=0}^\infty \frac{(-1)^n}{n!} = -e^{-1}\).
For the second part \(\sum_{k=1}^\infty \frac{2(-1)^{k+1}}{(k-1)!}\):
Let \(m = k-1\). As \(k \to 1, m \to 0\).
\(\sum_{m=0}^\infty \frac{2(-1)^{m+2}}{m!} = 2\sum_{m=0}^\infty \frac{(-1)^m}{m!} = 2e^{-1}\).
Total Sum = \(-e^{-1} + 2e^{-1} = e^{-1} = 1/e\).
Step 4: Final Answer:
The sum is \(1/e\).