$\limsup_{n\to \infty} \sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}\leq 1$?

Question

Let $(a_n)^i, i=1,2,3$, be complex sequences with $\limsup_{n\to \infty} \sqrt[n]{\lvert (a_n)^i\rvert}\leq 1, i=1,2,3$. I want to show that $$\limsup_{n\to \infty} \sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}\leq 1$$ Proof: Since $\limsup_{n\to \infty} \sqrt[n]{\lvert (a_n)^i\rvert}\leq 1, i=1,2,3$, for every $\epsilon>0$, there is a natural number $N_i$ such that, for every $n\geq N_i$, we have $\sqrt[n]{\lvert (a_n)^i\rvert}< 1+\epsilon \Leftrightarrow \lvert (a_n)^i\rvert < (1+\epsilon)^n$. Since $$\sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}\leq \sqrt[n]{\lvert (a_n)^1\rvert + \lvert (a_n)^2\rvert + \lvert (a_n)^3\rvert},$$ it follows for all $n\geq \max\{N_1,N_2,N_3\}$ that $$\sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}<\sqrt[n]{3(1+\epsilon)^n}=\sqrt[n]{3} (1+\epsilon)$$ Since the limit superior is the largest accumulation point of a sequence, it immediatly follows that $$\limsup_{n\to \infty} \sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}\leq 1$$ cause otherwise, we could make $\sqrt[n]{3}(1+\epsilon)$ as close to 1 as we want for all $n\geq N$ with $N$ sufficiently large which would contradict $$\limsup_{n\to \infty} \sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert} > 1$$

Answer 1

That is correct, but a little bit sloppy at the end. After $$ \sqrt[n]{\Vert ((a_n)^1, (a_n)^2, (a_n)^3 )\Vert}<\sqrt[n]{3} (1+\epsilon) $$ for $n\geq \max\{N_1,N_2,N_3\}$ I would continue with $$ \limsup_{n \to \infty} \sqrt[n]{\Vert ((a_n)^1, (a_n)^2, (a_n)^3 )\Vert} \le \limsup_{n \to \infty} \sqrt[n]{3} (1+\epsilon) \\ = \lim_{n \to \infty} \sqrt[n]{3} (1+\epsilon) = 1 + \epsilon $$ and since this holds for all $\epsilon > 0$ the desired conclusion $$ \limsup_{n \to \infty} \sqrt[n]{\Vert ((a_n)^1, (a_n)^2, (a_n)^3 )\Vert} \le 1 $$ follows.

(In other words, consider first what happens for $n\to \infty$, and then what happens for $\epsilon \to 0$.)