Show that $\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} = \max\{c_1,c_2,\ldots,c_m\}$ 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?
 A: 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}
A: 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 $\lim_{n\to\infty} \sqrt[n]{m} = 1$.
A: 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.
