$\lim_{{n}\to {\infty}}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=?$ Find the limit of the following sequence:
$$\lim_{{n}\to {\infty}}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=?$$
I've tried using the Stolz-Cesaro theorem. I've set $a_n=1^p+2^p+\cdots+n^p$, and $b_n=n^{p+1}$.
After some transformations I got $$\lim_{{n}\to{\infty}} \frac{(n+1)^p}{(n+1)^p-n^p},$$ but after expanding the binomials, I could not make any transformations that would give me a satisfactory result. Can you help me please? Thank you very much.
 A: Hint: $S(n,p) = \dfrac{1}{n}\left(\left(\dfrac{1}{n}\right)^p + \left(\dfrac{2}{n}\right)^p+\cdots \left(\dfrac{n}{n}\right)^p\right)\to \displaystyle \int_{0}^1 x^pdx$
A: Applying Stolz-Cesaro leads to:
$$\lim_{{n}\to{\infty}} \frac{(n+1)^p}{(n+1)^{p+1}-n^{p+1}} = \lim_{{n}\to{\infty}} \frac{1}{n\left(1+\frac{1}{n}\right)^{p+1}-n}$$
dividing numerator and denominator by $n^{p}$.
Expand the denominator in power series:
$$n\left(1+\frac{1}{n}\right)^{p+1}-n = \binom{p+1}{1} + \binom{p+1}{2}\frac{1}{n}+\mathcal{O}(1/n^2)$$
Clearly it approaches $p+1$ as $n \to \infty$.
Hence, your required limit is $\frac{1}{p+1}$
If you are looking for slightly different arguments apart from the Stolz-Cesaro or the already mentioned idea in the other answer,
Here's a nice one by @Jack D'Aurizio
$\displaystyle k(k-1)\cdot\ldots\cdot(k-p+1)=p!\binom{k}{p}\leq k^p \leq p!\binom{k+p}{p}=(k+p)\cdot\ldots\cdot(k+1)$
for each $1 \le k \le n$
Thus,
$$p!\binom{n+1}{p+1} = p!\sum\limits_{k=1}^{n}\binom{k}{p}\leq \sum\limits_{k=1}^{n} k^p \leq p!\sum\limits_{k=1}^{n}\binom{k+p}{p} = p!\binom{n+p+1}{p+1}$$
Thus, $\displaystyle \lim_{{n}\to {\infty}}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}= \frac{1}{p+1}$ by Squeeze Theorem.
