Question

Value of : \(\sum_{k=1}^{\infty} \frac{(-1)^k \cdot k(k+1)}{k!}\)

• A. \(-\frac{1}{e}\)
• B. \(-\frac{2}{e}\)
• C. \(-\frac{3}{e}\)
• D. \(-\frac{4}{e}\)

Hint:

For terms like \( \frac{poly(k)}{k!} \), always resolve the polynomial into terms of falling factorials like \( k, k(k-1), k(k-1)(k-2) \) etc.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The series involves the expansion of \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
For \( x = -1 \), we have \( e^{-1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \).
Step 2: Key Formula or Approach:
We simplify the numerator \( k(k+1) = k(k-1) + 2k \).
The general term is \( t_k = \frac{(-1)^k [k(k-1) + 2k]}{k!} \).
Step 3: Detailed Explanation:
\[ S = \sum_{k=2}^{\infty} \frac{(-1)^k k(k-1)}{k(k-1)(k-2)!} + \sum_{k=1}^{\infty} \frac{2k(-1)^k}{k(k-1)!} \] \[ S = \sum_{k=2}^{\infty} \frac{(-1)^k}{(k-2)!} + 2 \sum_{k=1}^{\infty} \frac{(-1)^k}{(k-1)!} \] In the first sum, let \( m = k-2 \):
\[ S_1 = \sum_{m=0}^{\infty} \frac{(-1)^{m+2}}{m!} = (-1)^2 \sum_{m=0}^{\infty} \frac{(-1)^m}{m!} = e^{-1} \] In the second sum, let \( n = k-1 \):
\[ S_2 = 2 \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n!} = -2 \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = -2e^{-1} \] Total Sum \( S = e^{-1} - 2e^{-1} = -e^{-1} = -\frac{1}{e} \).
Step 4: Final Answer:
The value of the series is \( -\frac{1}{e} \).