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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.

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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*}

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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.

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

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