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Let $m\in \mathbb{N}$ and $c_1,c_2,\ldots,c_m \in \mathbb{R}_+$. Show that $$\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} = \max\{c_1,c_2,\ldots,c_m\}$$

My attempt: Since $$\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} \leq \lim_{n\rightarrow \infty}\sqrt[n]{\max\{c_1,c_2,\ldots,c_m\}} = \lim_{n\rightarrow \infty} \sqrt[n]{n}\sqrt[n]{\max\{c_1,c_2,\ldots,c_m\}}=\lim_{n \rightarrow \infty} \max\{\sqrt[n]{c_1^n},\sqrt[n]{c_2^n},\ldots,\sqrt[n]{c_m^n}\}=\lim_{n \rightarrow \infty}\max\{c_1,c_2,\ldots,c_m\}=\max\{c_1,c_2,\ldots,c_m\}$$

it follows that $\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n}$ is bounded, but I don't think it's monotonically decreasing, at least I can't prove this. Can anybody tell me whether the approach I have chosen is a good one, whether what I have done is correct and how to finish the proof?

Martin Sleziak
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eager2learn
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  • Not sure why you'd think that first inequality is true. Maybe missing something? – Thomas Andrews May 24 '14 at 13:15
  • One thing to note is: You don't have a $\lim{\sqrt[n]{c_1^n\cdots}}$ until you prove that the limit exists. So any statement that you wrote with the limit before actually proving it's existence is not true. – hrkrshnn May 24 '14 at 13:16
  • Your upper bound proof looks correct assuming that your second and third steps were supposed to be: $\lim_{n\rightarrow \infty}\sqrt[n]{n\max{c_1^n,c_2^n,\ldots,c_m^n}} = \lim_{n\rightarrow \infty} \sqrt[n]{n}\sqrt[n]{\max{c_1^n,c_2^n,\ldots,c_m^n}}$ – Peter Woolfitt May 24 '14 at 13:19
  • @boywholived $m$ is finite, so that should be fine? – Peter Woolfitt May 24 '14 at 13:21
  • @PeterWoolfitt. Oh yes, I just meant to convey an point and was lazy to type. Sorry if that caused a confusion. – hrkrshnn May 24 '14 at 13:22

3 Answers3

18

The short proof.

Let $c=\max\{c_1,c_2,\dots,c_m\}$ and note that:

$$c^n \leq c_1^n+c_2^n+\dots+c_m^n \leq mc^n$$

Now take the $n$th root, and see that $\lim_{n\to\infty} \sqrt[n]{m} = 1$.

Thomas Andrews
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2

Being $\{c_1,\cdots,c_m \}$ a fine set of number, exist and index $i\in [1,m]$ such that $$ \max\{c_1,\cdots,c_m \}=c_i. $$ Then results $$ \sqrt{c_1^n+\cdots+c_m^n} = \sqrt{c_i ^ n \Big(\frac{c_1^n}{c_i^n}+\frac{c_2^n}{c_i^n}+\frac{c_3^n}{c_i^n}+\cdots+1+\cdots+\frac{c_m^n}{c_i^n}\Big)} $$ where the number $1$ correpond to the $i$-th element. Being $c_i$ the max, then every ratio $c_1/c_i ,c_2/c_i,\cdots,c_m/c_i \leq 1$, i.e $$ \frac{c_j}{c_i}< 1,\,\forall j=1,\cdots,m,\, j\neq i $$ then for $n\to \infty$ follows $$ \Big(\frac{c_j}{c_i}\Big)^n\to 0,\, \forall j\neq i $$ So being $c_1>0$ then \begin{eqnarray} \lim_{n\to \infty}{\sqrt[1/n]{c_1^n+\cdots+c_m^n}}&=&\lim_{n\to\infty}{\sqrt{c_i ^ n \Big(\frac{c_1^n}{c_i^n}+\frac{c_2^n}{c_i^n}+\frac{c_3^n}{c_i^n}+\cdots+1+\cdots+\frac{c_m^n}{c_i^n}\Big)}}\\ &=&\lim_{n\to\infty}{c_i\sqrt{\frac{c_1^n}{c_i^n}+\frac{c_2^n}{c_i^n}+\frac{c_3^n}{c_i^n}+\cdots+1+\cdots+\frac{c_m^n}{c_i^n}}}\\ &=&\lim_{n\to\infty}{c_i\sqrt{\Big(\frac{c_1}{c_i}\Big)^n+\cdots+1+\cdots+\Big(\frac{c_m}{c_i}\Big)^n}}\\ &=&\lim_{n\to\infty}{c_i \cdot \sqrt{1}}\\ &=&c_i=\max\{c_1,\cdots,c_m\}. \end{eqnarray}

Gio712
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You can see there is an error in your approach if you consider a simple example. Let $c_1=2$ and $c_2=\cdots=c_m=0$. Then

$$\lim_{n\to\infty}\sqrt[n]{c_1^n+c_2^n+\cdots+c_m^n}=\lim_{n\to\infty}\sqrt[n]{2^n}=2$$ but

$$\lim_{n\to\infty}\sqrt[n]{\max\{c_1,c_2\ldots,c_m\}}=\lim_{n\to\infty}\sqrt[n]{2}=1$$

so your first inequality does not always hold.

Barry Cipra
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