Question

Evaluate: \[ \lim_{n \to \infty} \frac{1}{n^{k+1}} \left[ 2^k + 4^k + 6^k + \dots + (2n)^k \right]. \]

• A. \( \frac{2^k}{k} \)
• B. \( \frac{2^{k+1}}{k+1} \)
• C. \( \frac{2^k}{k+1} \)
• D. \( \frac{2^k}{k-1} \)

Answer 1

The Correct Option is C

1. The sum is:

\( S_n = 2^k + 4^k + 6^k + \cdots + (2n)^k \).

Factor out \( 2^k \):

\( S_n = 2^k \left[ 1^k + 2^k + 3^k + \cdots + n^k \right] \).

2. The term inside the brackets is the \( k \)-th power sum:

\[ \sum_{r=1}^n r^k \sim \frac{n^{k+1}}{k+1}, \text{ as } n \to \infty. \]

3. Substitute:

\( S_n \sim 2^k \cdot \frac{n^{k+1}}{k+1}. \)

4. Divide \( S_n \) by \( n^{k+1} \):

\[ \frac{S_n}{n^{k+1}} = \frac{2^k}{k+1}. \]

5. Taking the limit as \( n \to \infty \):

\[ \lim_{n \to \infty} \frac{1}{n^{k+1}} \left[ 2^k + 4^k + 6^k + \cdots + (2n)^k \right] = \frac{2^k}{k+1}. \]

Thus, the correct answer is \( \frac{2^k}{k+1} \).