How to evaluate: $\lim\limits_{n\to\infty} \frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$ How to evaluate: $$\lim_{n\to\infty} \dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$$
when 
$i)$ $p\in\mathbb R,p\neq0$
$ii)\space p=0$
So for $i)$ I tried using Stolz–Cesàro theorem and Binomial theorem and If I didn't mess up I got $1$. But I'm unsure about it, but for $ii)$ and I don't have a clue where to begin with.

We can apply the Stolz-Cesàro theorem for positive real values $p,  p\ne  1$
and consider other values  of $p$ separately.
We consider for $p\in\mathbb{R}$:
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}\tag{1}
\end{align*}

Case $p>1, 0<p<1$:
If $p>1$ resp. $0<p<1$ the sequence $(n^p)_{n\geq 1}$ is strictly monotone increasing and unbounded. We can apply the Stolz-Cesàro theorem by letting
\begin{align*}
a_n&=1^{p-1}+2^{p-1}+\cdots+n^{p-1}\\
b_n&=n^p=\underbrace{n^{p-1}+n^{p-1}+\cdots+n^{p-1}}_{n\ \mathrm{ times}}
\end{align*}
Since
\begin{align*}
\lim_{n\to \infty}\frac{a_n-a_{n-1}}{b_n-b_{n-1}}&=\lim_{n\to\infty}\frac{n^{p-1}}{n^p-(n-1)^{p}}\\
&=\lim_{n\to\infty}\frac{n^{p-1}}{n^p-(n^p-pn^{p-1}+\binom{p}{2}n^{p-2}-\cdots)}\\
&=\lim_{n\to\infty}\frac{n^{p-1}}{pn^{p-1}-\binom{p}{2}n^{p-2}+\cdots}\\
&=\frac{1}{p}
\end{align*}
we have according to the theorem
\begin{align*}
\lim_{n\to \infty}\frac{a_n}{b_n}&=\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}\color{blue}{=\frac{1}{p}}
\end{align*}

Case $p=1$:
We obtain
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}
=\lim_{n\to\infty}\frac{\overbrace{1+1+\cdots+1}^{n\mathrm{\ times}}}{n}=\lim_{n\to\infty}\frac{n}{n}
\color{blue}{=1}
\end{align*}

Case $p=0$:
We obtain
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}
=\lim_{n\to\infty}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)=\sum_{n=1}^\infty \frac{1}{n}\color{blue}{=\infty}
\end{align*}
since the harmonic series is divergent.

Case $p<0$:
We set $q:=-p$ and obtain with $q>0$:
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}
=\lim_{n\to \infty}n^q\left(1+\frac{1}{2^{q+1}}+\cdots+\frac{1}{n^{q+1}}\right)\geq \lim_{n\to\infty} n^q
\color{blue}{=\infty}
\end{align*}

We summarize:
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}=
\begin{cases}
\frac{1}{p}&p>0\\
\infty&p\leq 0
\end{cases}
\end{align*}

