# 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.

• I would use series / integral comparison to bound below and above $1^{p-1}+2^{p-1}+…+n^{p-1}$. – mathcounterexamples.net Jan 25 at 10:33
• – lab bhattacharjee Jan 25 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 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 at 10:40
• – Martin Sleziak Feb 3 at 11:21

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:

If $$p>1$$ resp. $$0 the sequence $$(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*}

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.

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