Evaluate $\displaystyle\lim_{n\to\infty} \dfrac{1^k + 2^k + \cdots + n^k}{n^{k+1}}$ for $k > -1$:
- $0$
- $\dfrac{1}{k+1}$
- $1$
- $k$
Show answer and solution
Correct answer: B. $\dfrac{1}{k+1}$
Recognise as Riemann sum: $\sum (r/n)^k \cdot (1/n) \to \int_0^1 x^k\, dx = 1/(k+1)$.