What is the result of $ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $ and why? What is the result of the next limit: 
$$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$
Why (theorem)?
 A: This is a Riemann sum, which in this limit takes on the value of an integral:
$$\lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n \frac{i^k}{n^k} = \int_0^1 dx \, x^k = \frac{1}{k+1}$$
so long as $k \gt -1$.
A: You can apply Lemma Stolz-Cesaro:
$$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}}= \lim_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1}$$
A: Just to clarify Ron's answer, the result you're looking for is the following:
$$\lim_{n \to \infty} \frac{K}{n}\sum_{i=1}^n f\left(\frac{Ki}{n}\right)=\int_0^K f(x) \, \mathrm{d}x.
$$
In your case, $f(x) = x^k$, and $K=1$. The trick here is that whenever you deal with the limit of a sum, factor outside the sum $1/n$ and see whether you can group the sum term to reduce to the Riemann sum expression. If you can, switch to integral form.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\lim_{n\ \to\ \infty}{\sum^{n}_{i\ =\ 1}i^{k} \over n^{k + 1}}:\ {\large ?}.
     \qquad k \in {\mathbb R}}$.



*

*
$\ds{\large\dsc{k < -1}:}$
\begin{align}&\lim_{n\ \to\ \infty}{\sum^{n}_{i\ =\ 1}i^{k} \over n^{k + 1}}
=\lim_{n\ \to\ \infty}\pars{%
n^{\verts{k + 1}}\sum^{n}_{i\ =\ 1}{1 \over i^{\verts{k}}}}
=\color{#66f}{\large\infty}
\end{align}
Note that
$\ds{\lim_{n\ \to\ \infty}\sum^{n}_{i\ =\ 1}{1 \over i^{\verts{k}}}=\zeta\pars{\verts{k}}}$


*
$\ds{\large \dsc{k = -1}:}$
\begin{align}&\lim_{n\ \to\ \infty}{\sum^{n}_{i\ =\ 1}i^{k} \over n^{k + 1}}
=\lim_{n\ \to\ \infty}\sum^{n}_{i\ =\ 1}{1 \over i}=\color{#66f}{\large\infty}
\end{align}
Note that
$\ds{\sum^{n}_{i\ =\ 1}{1 \over i}=H_{n}}$ where $\ds{H_{n}}$ is the Harmonic Number which diverges logarithmically
$\ds{\pars{~H_{n} \sim \ln\pars{n}\ \mbox{when}\ n \gg 1~}}$.


*
$\ds{\large \dsc{k > -1}:}$
\begin{align}\lim_{n\ \to\ \infty}{\sum^{n}_{i\ =\ 1}i^{k} \over n^{k + 1}}
&=\lim_{n\ \to\ \infty}
{\sum^{n + 1}_{i\ =\ 1}i^{k} - \sum^{n}_{i\ =\ 1}i^{k}\over
\pars{n + 1}^{k + 1} - n^{k + 1}}
=\lim_{n\ \to\ \infty}
{\pars{n + 1}^{k} \over \pars{n + 1}^{k + 1} - n^{k + 1}}
=\color{#66f}{\large{1 \over k + 1}}
\end{align}

