For a bounded sequence prove $\text{lim}_{n \to \infty} \left(\sum\limits_{k=1}^n|a_k|^n \right)^{1/n}=\text{sup}_{k \in \mathbb{N}}|a_k|$ I want to show that $$\text{lim}_{n \to \infty} \left(\sum\limits_{k=1}^n|a_k|^n \right)^{1/n}=\text{sup}_{k \in \mathbb{N}}|a_k|$$ given a bounded sequence $(a_k)$. I think its best to do that by the epsilon method, but I seem to oversee something. Let $\epsilon>0$. Since the sequence is bounded we have that for all $\epsilon'>0$ $$a-\epsilon'<|a_t| <\left(\sum\limits_{k=1}^n|a_k|^n \right)^{1/n},$$ for $n \geq t$ where $a:= \text{sup}_{k \in \mathbb{N}} |a_k|.$ Hence $-\epsilon<\left(\sum\limits_{k=1}^n|a_k|^n \right)^{1/n}-a$. Unfortunately I don't know how to go from here or whether what I have tried is even expedient. Any advice is greatly appreciated, thank you in advance.
 A: Here is a start: Let $M=\sup |a_k|.$ Then
$$ \left(\sum_{k=1}^n|a_k|^n \right)^{1/n} \le \left(\sum_{k=1}^nM^n \right)^{1/n}= \left( nM^n \right)^{1/n} = n^{1/n}M.$$
Recall that as $n\to \infty,$ $n^{1/n}\to 1.$
A: HINT
We have that
$$\left(|a_{sup}|^n\right)^\frac1n\le \left(\sum_{k=1}^n{|a_k|^n}\right)^\frac1n\le\left(\sum_{k=1}^n{|a_{sup}|^n}\right)^\frac1n$$
with $a_{sup}=\text{sup}_{k \in \mathbb{N}}|a_k|$
A: Since it is bounded then there is some positive number
$$
M = \sup\{|a_i|\}_{i=1}^\infty 
$$
Assuming that this values is attained for (at least) some $i = m$ (which is not generally the case), then by dividing and multiplying thegiven sequence by $M$ we get
\begin{align}
\left(\sum_{k=1}^n|a_k|^n\right)^{\frac{1}{n}} &= (|a_1|^n+...+|a_n|^n)^\frac{1}{n}\\
&= M\left((\frac{|a_1|}{M})^n+\dots+(\frac{|a_m|}{M})^n+\dots+(\frac{|a_n|}{M})^n\right)^{\frac{1}{n}}\\
&=M\left((\frac{|a_1|}{M})^n+\dots+1+\dots+(\frac{|a_n|}{M})^n\right)^{\frac{1}{n}} \ge M
\end{align}
The other part of the answer is in @zhw. answer
