Prove that $\lim_{n\rightarrow\infty}\frac{1}{n^{p+1}}\sum_{k=1}^{n}k^{p}=\frac{1}{p+1} $ 
I need to prove that
$$
\lim_{n\to\infty}\frac{1}{n^{p+1}}\sum_{k=1}^n k^p
 = \frac{1}{p+1}
$$

By Stulz lemma, $$\frac{(n+1)^{p}}{(n+1)^{p+1}-n^{p+1}}=\lim_{n\rightarrow\infty}\frac{1}{n^{p+1}}\sum_{k=1}^{n}k^{p}=\frac{1}{p+1}$$
and
$$\frac{(n+1)^{p}}{(n+1)^{p+1}-n^{p+1}}\geq\frac{1}{\frac{p+1+\frac{1}{n}}{1+\frac{1}{n}}}\rightarrow  \frac{1}{1+p}$$
And know I'm stuck..
 A: Hint:
$$\int_1^nx^p dx \leq \sum_{k=1}^n k^p \leq \int_1^{n+1}x^p dx$$
Precalculus:
Use induction. For the upper bound assume that
$$\sum_{k=1}^n k^p \leq \frac{(n+1)^{p+1}}{p+1}$$
holds. Then show that $(n+1)^p + \frac{(n+1)^{p+1}}{p+1} \leq \frac{(n+2)^{p+1}}{p+1}$ to conclude that $\sum_{k=1}^{n+1} k^p \leq \frac{(n+2)^{p+1}}{p+1}$ and since it holds for $n=1$ it follows by induction that the inequality hold for all $n$. For the lower bound assume that
$$\frac{n^{p+1}-1}{p+1}\leq \sum_{k=1}^n k^p$$
holds. Then show that  $\frac{n^{p+1}-1}{p+1} + (n+1)^p\geq \frac{(n+1)^{p+1}}{p+1}$ and conclude.
A: We have:
$$ k(k-1)\cdot\ldots\cdot(k-p+1)=p!\binom{k}{p}\leq k^p \leq p!\binom{k+p}{p}=(k+p)\cdot\ldots\cdot(k+1). $$
Summing over $k$:
$$ p!\binom{n+1}{p+1}\leq\sum_{k=1}^{n}k^p\leq p!\binom{n+p+1}{p+1} $$
hence:
$$ \lim_{n\to +\infty}\frac{1}{n^{p+1}}\sum_{k=1}^{n}k^p = \frac{1}{p+1}$$
follows by squeezing.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\lim_{n\ \to\ \infty}{1 \over n^{p + 1}}\sum_{k\ =\ 1}^{n}k^{p}
    ={1 \over p + 1}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large%
\lim_{n\ \to\ \infty}{1 \over n^{p + 1}}\sum_{k\ =\ 1}^{n}k^{p}}
=\lim_{n\ \to\ \infty}{1 \over n}\sum_{k\ =\ 1}^{n}\pars{k \over n}^{p}
=\int_{0}^{1}x^{p}\,\dd x
=\left.{x^{p + 1} \over p + 1}\right\vert_{\, x\ =\ 0}^{\,x\ =\ 1}
=\color{#66f}{\large{1 \over p + 1}}
\end{align}

A: Use the Faulhaber formula:
$$\sum_{k=1}^n k^p = \dfrac1{p+1} \sum_{j=0}^p (-1)^j \dbinom{p+1}j B_j n^{p+1-j}$$
where $B_j$ are the Bernoulli numbers.
