Question

\( \lim_{n \to \infty} \left( \frac{1}{1 + n^5} + \frac{2^4}{2^5 + n^5} + \frac{3^4}{3^5 + n^5} + \ldots + \frac{n^4}{n^5 + n^5} \right) = \)

• A. \( \frac{1}{5} \log 3 \)
• B. \( \frac{1}{3} \log 5 \)
• C. \( \frac{1}{2} \log 5 \)
• D. \( \log \sqrt{2} \)

Hint:

Convert Riemann sum to a definite integral using \( x = \frac{r}{n} \), then apply substitution if needed.

Answer 1

The Correct Option is D

We are given: \[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r^4}{r^5 + n^5} = \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{(r/n)^4}{(r/n)^5 + 1} \] Let \( x = \frac{r}{n} \), then as \( n \to \infty \), this becomes: \[ \int_0^1 \frac{x^4}{x^5 + 1} dx \] Substitute \( x^5 = t \Rightarrow x^4 dx = \frac{dt}{5} \).
Limits: \( t = 0 \) to \( t = 1 \), so: \[ \int_0^1 \frac{1}{x^5 + 1} x^4 dx = \frac{1}{5} \int_0^1 \frac{1}{t + 1} dt = \frac{1}{5} \log(1 + t) \Big|_0^1 = \frac{1}{5} \log 2 \] So, answer is: \( \log \sqrt{2} = \frac{1}{2} \log 2 \)