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.
 A: As $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx,$$
we have $$\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(\frac rn\right)^a=\int_0^1x^a\ dx$$
A: You can use L'hopital's rule as 

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

A: It also follows from the Cesaro theorem.
A: Look at the:
$$\sum_{r=1}^nr^a=1^a+2^a+\dotsb+n^a$$
part. It turns out that this is a polynomial, whose leading term is $\dfrac1{a+1}r^{a+1}$. For example:
\begin{align}
\sum_{r=1}^nr^1&=\frac12n^2+\frac12n\\
\sum_{r=1}^nr^2&=\frac13n^3+\frac12n^2+\frac16n\\
\sum_{r=1}^nr^3&=\frac14n^4+\frac12n^3+\frac14n^2\\
\sum_{r=1}^nr^4&=\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n\\
\sum_{r=1}^nr^5&=\frac16n^6+\frac12n^5+\frac5{12}n^4-\frac1{12}n^2
\end{align}
How to prove?
One thing to note is that if:
$$\sum_{r=1}^nf(r)=S(r)$$
then:
$$S(r)-S(r-1)=f(r)$$
(prove this!). Also note that $x^{a+1}-(x-1)^{a+1}$ is a polynomial whose first term is $(a+1)x^a$ (prove this too!). Can you use these facts to prove the statement in bold above? (Once you have this, it's easy to answer the stated problem.)
