Proving $\lim_{n \rightarrow \infty} \frac{\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$ How do we prove that $$\lim_{n \rightarrow \infty} \dfrac{\displaystyle\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$$?
This type of terms appear in problems on limits, but I am unable to prove this.
Please help me out.

You can use L'hopital's rule as 

$$ \lim_{n \rightarrow \infty} \dfrac{\displaystyle\sum_{r=1}^{n} r^a}{n^{a+1}} = \lim_{n \rightarrow \infty} \dfrac{\displaystyle n^a}{(1+a)n^{a}} = \frac{1}{a+1}$$

