How to compute the limit $\lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n}$? Let $a_1, \ldots, a_n > 0$. How to compute the limit $\lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n}$?
My solution: 
$$
\lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n} \\
= \lim_{n \to \infty} (1 + a_1^n + \cdots + a_m^n - 1)^{1/n} \\
= \lim_{n \to \infty} \left( (1 + a_1^n + \cdots + a_m^n - 1)^{\frac{1}{a_1^n + \cdots + a_m^n - 1}} \right)^{ \frac{a_1^n + \cdots a_m^n-1}{n}}.
$$
But then we need to have the condition $a_1^n + \cdots + a_m^n - 1 \to 0$ ($n \to \infty$). 
Thank you very much.
 A: Let $A = \max \{a_1, \cdots, a_m\}$. Then
$$\left( a_1^n + \cdots + a_m^n \right)^{\frac 1n} \ge (A^n)^\frac{1}{n} = A.$$
On the other hand,
$$\left( a_1^n + \cdots + a_m^n \right)^{\frac 1n}\le (A^n + A^n + \cdots + A^n)^{\frac 1n} = m^{\frac 1n} A$$
This implies 
$$\liminf \left( a_1^n + \cdots + a_m^n \right)^{\frac 1n} \le A.$$
So the limit is $A$. 
Remark: Note that 
$$\left( a_1^n + \cdots + a_m^n \right)^{\frac 1n}$$
is the $n$-th norm of the vector $v =(a_1, a_2, \cdots, a_m)$. The above shows that 
$$ \lim_{n \to \infty} \|v\|_n = \| v\|_\infty = \max\{ a_1, \cdots, a_m\}.$$
This has a generalization in measure theory, concerning the limit of the $L^p$ norm of a function $f$.
A: Think about the part under the root. The largest term will become more and more significant compared to the others. In the limit it will completely dominate the other terms. The root effectively undoes the power so the function is equivalent to the maximum function.
$$\lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n}=\max(a_1,a_2,\cdots,a_m)$$
