Proving that the limit of $\root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n}$ is $\max(a_1,...a_k)$

Question

Prove that: $$\lim_{n \to \infty} \root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n} = \max \left\{ {{a_1},{a_2}...{a_k}} \right\}$$

I am familiar with the theorem which says that if $$\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{a_{n - 1}}}} = L$$

then,
$$\mathop {\lim }\limits_{n \to \infty } \root n \of {{a_n}} = L$$

So, I tried evaluating the expreesion:
$${{{a_1}^n + {a_2}^n + ... + {a_k}^n} \over {{a_1}^{n - 1} + {a_2}^{n - 1} + ... + {a_k}^{n - 1}}}$$

but pretty much got stuck here. Is this the right path to go?

Answer 1

$$\sqrt[n]{a_{\max}^n} \leq \sqrt[n]{a^n_1 + \dots + a_k^n} \leq \sqrt[n]{k \cdot a_{\max}^n}$$

Answer 2

Let's suppose, without any loss of generality, that $a_1 = \max \left\{ {{a_1},{a_2}...{a_k}} \right\}$. Then:

$$ \root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n} = \root n \of {a_1^n \left (1+\left (\dfrac{a_2}{a_1} \right )^n + ... + \left (\dfrac{a_k}{a_1} \right )^n \right )} \leq \root n \of {{a_1}^n \cdot k} = a_1 \root n \of {k}$$

Since $\dfrac{a_i}{a_1} \leq 1$, $i=1, ..., k$.

Therefore, $\lim_{n \to \infty} \root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n} = a_1$

Answer 3

Although your question is already answered, notice that (for positive $(a_k)_{k\in \mathbb N} $) what you want to prove is equivalent to proving that $$||a_k||_p \to ||a_k||_\infty$$ as $p \to \infty$ (where $p$ is denoted with $n$ in your case), i.e. that the $l^p$-norm converges to the $l^\infty$-norm (for sequences). A complete proof can be found here.