Using one limit to compute other I've calculated $\lim_{n\to\infty}\dfrac{1^p+2^p+\cdots+n^p}{n^{p+1}}=\dfrac1{p+1}$ where $p\in\mathbb{N}$ fixed. I feel it should help me get this one $\lim_{n\to\infty}\left(\dfrac{1^p+2^p+\cdots+n^p}{n^{p}}-\dfrac{n}{p+1}\right)$, but I'm not sure how. Any hints?
 A: I don't know how to get the second limit by using the first one since $$\lim_{n\to\infty}\left(\dfrac{1^p+2^p+\ldots+n^p}{n^{p}}-\dfrac{n}{p+1}\right) = \lim_{n\to\infty}n\left(\dfrac{1^p+2^p+\ldots+n^p}{n^{p+1}}-\dfrac{1}{p+1}\right)$$
which is of $0\times\infty$ type.
But both limits can be computed using  Stolz–Cesàro theorem through binomial formula
Added:
\begin{align}
&\lim_{n\to\infty}\left(\dfrac{1^p+2^p+\ldots+n^p}{n^{p}}-\dfrac{n}{p+1}\right)\\
=&\lim_{n\to\infty}\left(\dfrac{(p+1)(1^p+2^p+\ldots+n^p)-n^{p+1}}{(p+1)n^{p}}\right) \\
=&\lim_{n\to\infty}\left(\dfrac{(p+1)(n+1)^p-(n+1)^{p+1}+n^{p+1}}{(p+1)((n+1)^{p}-n^p)}\right)\\
=&\lim_{n\to\infty}\left(\dfrac{\frac{(p+1)p}{2}n^{p-1}+\text{term of lower order}}{(p+1)pn^{p-1} + \text{term of lower order}}\right)\\
=& \frac{1}{2}
\end{align}
In the comment, @Hamou gives a way to get directly the limit by application of theorem on convergence rate of Riemann sum. For exact statement of the theorem, look into the reference therein.
A: Hint: Use the equivalence $1^p+2^p+\ldots+n^p\sim \dfrac{n^{p+1}}{p+1}$.
$$\dfrac{\sum_{k=1}^nk^p}{n^{p+1}}=\frac{1}{n}\sum_{k=1}^n\left(\dfrac{k}{n}\right)^p=\frac{1-0}{n}\sum_{k=1}^n\left(0+\dfrac{k}{n}(1-0)\right)^p\to\int_0^1x^pdx=\dfrac{1}{p+1}$$
