Can we prove that $\lim_{n\to\infty} \frac{1^{n_0}+2^{n_0}+\cdots+n^{n_0}}{n^{n_0+1}}$ is finite for any $n_0\in\Bbb N$ without a direct computation? 
Can we prove without direct calculation that this limit is finite for any natural number $n_0 \in \mathbb{N}$?

$$ \lim_{n \to \infty} \frac{1^{n_0}+2^{n_0}+\cdots+n^{n_0}}{n^{n_0+1}} $$ 
 A: The numerator can be expressed as a polynomial $P(n)$ of degree $n_0+1$, because $P(n)-P(n-1)=n^{n_0}$ is a polynomial of degree $n_0$.
So the limit of $$\dfrac{P(n)}{n^{n_0+1}}$$ is finite.

By the Faulhaber's formulas, the limit is $\dfrac1{n_0+1}$.
A: For positive integer  $k\leq n$ we have    $$\int_{k-1}^kx^{n_0}dx<k^{n_0}<\int_k^{k+1}x^{n_0}dx.$$ Now add from $k=1$ to $k=n.$
A: Alternative method: Cesàro-stolz theorem [if you've learned]. 
A: I'll use $p$ instead of $n_0$. The number you're interested in is:
$$
L_p=\lim_{n\to\infty}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}
$$
Alternatively, by factoring out $1/n$, we may write:
$$
L_p=\lim_{n\to\infty}\frac 1n\sum_{k=1}^n\left(\frac kn\right)^p
$$
The term in the limit is an average of $n$ values all of which are in $[0,1]$, provided $p>0$. The limit function itself then must be bounded to this interval.
This does not show that the limit exists, but provided it does, it must be in the unit interval.
A: Here's a completely elementary proof
that just uses
Bernoulli's inequality.
$\begin{array}\\
(x+1)^m-x^m
&=x^m((1+1/x)^m-1)\\
&\ge x^m(1+m/x-1)
\qquad\text{by Bernoulli}\\
&=mx^{m-1}\\
\end{array}
$
Therefore
$\sum_{k=0}^{n-1} k^{m-1}
\le \sum_{k=0}^{n-1}\frac1{m}((k+1)^m-k^m)
=\frac1{m}n^m
$
so,
for $m \ge 2$,
$\sum_{k=1}^{n} k^{m-1}
\le n^{m-1}+\frac1{m}n^m
$
or
$\frac1{n^m}\sum_{k=1}^{n} k^{m-1}
\le \frac1{n^m}(n^{m-1}+\frac1{m}n^m)
=\frac1{n}+\frac1{m}
$
which is bounded.
You have to work a little harder
to show that
$\frac1{m}$
is the actual limit.
A: As in Falling and rising factorials, define (using $p$ instead of $n_0$, as in a previous answer):
\begin{align*}
(k)_p & = k(k - 1)\cdots(k - p + 1), \\
k^{(p)} & = k(k + 1)\cdots(k + p - 1).
\end{align*}
Then:
\begin{align*}
(k + 1)_{p + 1} - k_{p + 1} & = (p + 1)(k)_p, \\
k^{(p + 1)} - (k - 1)^{(p + 1)} & = (p + 1)k^{(p)}.
\end{align*}
For every positive integer $k$,
$$
(k)_p \leqslant k^p \leqslant k^{(p)}.
$$
Hence, for every positive integer $n$,
$$
\frac{(n + 1)_{p + 1}}{p + 1} \leqslant \sum_{k=1}^n k^p \leqslant \frac{n^{(p + 1)}}{p + 1}.
$$
But
$$
\frac{(n + 1)_{p + 1}}{n^{p + 1}} \to 1 \text{ as } n \to \infty, \text{ and }
\frac{n^{(p + 1)}}{n^{p + 1}} \to 1 \text{ as } n \to \infty,
$$
therefore
$$
\frac{\sum_{k=1}^n k^p}{n^{p + 1}} \to \frac{1}{p + 1} \text{ as } n \to \infty.
$$
