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?

Question

Can we prove without direct calculation that this limit is finite for any natural number $n_0 \in \mathbb{N}$?

$$ \lim_{n \to \infty} \frac{1^{n_0}+2^{n_0}+\cdots+n^{n_0}}{n^{n_0+1}} $$

Answer 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 $$\dfrac{P(n)}{n^{n_0+1}}$$ is finite.

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By the Faulhaber's formulas, the limit is $\dfrac1{n_0+1}$.

Answer 2

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

Answer 3

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.

Answer 4

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 $k=n.$

Answer 5

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.

Answer 6

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 $$ \frac{\sum_{k=1}^n k^p}{n^{p + 1}} \to \frac{1}{p + 1} \text{ as } n \to \infty. $$