How to evaluate: $\lim\limits_{n\to\infty} \frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$

Question

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.

I would use series / integral comparison to bound below and above $1^{p-1}+2^{p-1}+…+n^{p-1}$. — mathcounterexamples.net, Jan 25 '19 at 10:33
Use https://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2/469886#469886 — lab bhattacharjee, Jan 25 '19 at 10:34
$$\lim_{n\to\infty}\frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}=\lim_{n\to\infty}\sum_{j=1}^n\left(\frac jn\right)^{p-1}\frac1n=\int_0^1x^{p-1}dx=\begin{cases}\frac1p,&p\ne0\\infty,&p=0\end{cases}$$ — Shubham Johri, Jan 25 '19 at 10:37
For $p=0$, the limit is the sum of the Harmonic series ${\frac1n}$. Proofs of its divergence are quite popular; see here — Shubham Johri, Jan 25 '19 at 10:40
For $p>0$ you can have a look at Evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}}$ (and the questions linked there) or 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? (and the questions linked there). — Martin Sleziak, Feb 03 '19 at 11:21

score 7 · Answer 1 · answered Jan 25 '19 at 10:39

7

Hint:

Just write $$\dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p} = \frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^{p-1}$$ and handle it as the limit of a Riemann sum.

answered Jan 25 '19 at 10:39

trancelocation

32,243

Sorry but I haven't learnt about integration yet and I don't know how to do that :( – iggykimi Jan 25 '19 at 18:10

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

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*}

How to evaluate: $\lim\limits_{n\to\infty} \frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$

2 Answers2