Finding $\lim \limits_{n \to \infty}{1^k + 2^k + 3^k + \dots + n^k - {1 \over k+1} n^{k+1} \over n^k}$ I am trying to evaluate
$$\lim \limits_{n \to \infty}{1^k + 2^k + 3^k + \dots + n^k - {1 \over k+1} n^{k+1} \over n^k}$$
I have tried several techniques I've covered so far, however, all of them give me an illegal expression such as ${0 \over 0}$ or fail otherwise. I am aware of the existence of Faulhaber's formula but would prefer not to rely on it. Any ideas?
 A: We know from the trapezoidal integration formula that (for $f\in C^2([0,1])$)
$$
\frac1n\left(\frac12f(0)+\sum_{k=1}^{n-1}f(\frac kn)+\frac12f(1)\right)=\int_0^1 f(x)dx + O(\frac1{n^2})
$$
In the present problem with $f(x)=x^k$ this means that
$$
\frac{1^k+2^k+…+n^k}{n^k}=\frac1{2n^{k}}+\frac1{2}+n \int_0^1 x^k\,dx+O\bigl(\frac1n\bigr)
\\
=\frac1{2n^{k}}+\frac1{2}+\frac{n}{k+1}+O\bigl(\frac1n\bigr)
$$
from which the requested limit can be read off as $\frac12$.
A: Using Stolz–Cesàro theorem as commented by @AK47.
$L = \lim\limits_{n \rightarrow \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} \\ = \lim\limits_{n \rightarrow \infty} \frac{(n+1)^k - \frac{(n+1)^{k+1}}{k+1} + \frac{n^{k+1}}{k+1}}{(n+1)^k-n^k}\\ = \lim\limits_{n \rightarrow \infty} \frac{(k-n)(n+1)^k + n^{k+1}}{(k+1)((n+1)^k-n^k)} \\ = \lim\limits_{n \rightarrow \infty}\frac{k\sum\limits_{i=0}^{k}\begin{pmatrix}k \\ i \end{pmatrix}n^i - \sum\limits_{i=0}^{k}\begin{pmatrix}k \\ i \end{pmatrix}n^{i+1} + n^{k+1}}{(k+1)(\sum\limits_{i=0}^{k}\begin{pmatrix}k \\ i \end{pmatrix}n^i - n^k)} \\ = \lim\limits_{n \rightarrow \infty}\frac{k\sum\limits_{i=0}^{k-1}\begin{pmatrix}k \\ i \end{pmatrix}n^i - \sum\limits_{i=0}^{k-2}\begin{pmatrix}k \\ i \end{pmatrix}n^{i+1}}{(k+1)(\sum\limits_{i=0}^{k-1}\begin{pmatrix}k \\ i \end{pmatrix}n^i)}$
Now, divide numerator and denominator by $n^{k-1}$ and take the limit $n \rightarrow \infty$ to get the following, 
$= \frac{k\begin{pmatrix}k \\ k-1\end{pmatrix} - \begin{pmatrix}k \\ k-2\end{pmatrix}}{(k+1)\begin{pmatrix}k \\ k-1\end{pmatrix}} = \frac{1}{2}$
