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Can we prove without direct calculation that this limit is finite for any natural number $n_0 in mathbbN$?

$$ lim_n to infty frac1^n_0+2^n_0+cdots+n^n_0n^n_0+1 $$

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 $$dfracP(n)n^n_0+1$$ is finite.

By the Faulhaber's formulas, the limit is $dfrac1n_0+1$.

For positive integer $kleq n$ we have $$int_k-1^kx^n_0dx<k^n_0<int_k^k+1x^n_0dx.$$ Now add from $k=1$ to $k=n.$

I'll use $p$ instead of $n_0$. The number you're interested in is:
$$
L_p=lim_ntoinftyfrac1^p+2^p+cdots+n^pn^p+1
$$
Alternatively, by factoring out $1/n$, we may write:
$$
L_p=lim_ntoinftyfrac 1nsum_k=1^nleft(frac knright)^p
$$
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.

Here's a completely elementary proof
that just uses
Bernoulli's inequality.

$beginarray\
(x+1)^m-x^m
&=x^m((1+1/x)^m-1)\
&ge x^m(1+m/x-1)
qquadtextby Bernoulli\
&=mx^m-1\
endarray
$

Therefore
$sum_k=0^n-1 k^m-1
le sum_k=0^n-1frac1m((k+1)^m-k^m)
=frac1mn^m
$
so,
for $m ge 2$,
$sum_k=1^n k^m-1
le n^m-1+frac1mn^m
$
or
$frac1n^msum_k=1^n k^m-1
le frac1n^m(n^m-1+frac1mn^m)
=frac1n+frac1m
$
which is bounded.

You have to work a little harder
to show that
$frac1m$
is the actual limit.

Alternative method: Cesàro-stolz theorem [if you've learned].

As in Falling and rising factorials, define (using $p$ instead of $n_0$, as in a previous answer):
beginalign*
(k)_p & = k(k - 1)cdots(k - p + 1), \
k^(p) & = k(k + 1)cdots(k + p - 1).
endalign*
Then:
beginalign*
(k + 1)_p + 1 - k_p + 1 & = (p + 1)(k)_p, \
k^(p + 1) - (k - 1)^(p + 1) & = (p + 1)k^(p).
endalign*
For every positive integer $k$,
$$
(k)_p leqslant k^p leqslant k^(p).
$$
Hence, for every positive integer $n$,
$$
frac(n + 1)_p + 1p + 1 leqslant sum_k=1^n k^p leqslant fracn^(p + 1)p + 1.
$$
But
$$
frac(n + 1)_p + 1n^p + 1 to 1 text as n to infty, text and
fracn^(p + 1)n^p + 1 to 1 text as n to infty,
$$
therefore
$$
fracsum_k=1^n k^pn^p + 1 to frac1p + 1 text as n to infty.
$$