Is $\limsup\sqrt[n]{|a_{n+1}|}=\limsup\sqrt[n]{|a_n|}$? Let $\{a_n\}_{n=1}^\infty$ be an arbitrary sequence of complex numbers. Does the equality
\begin{equation}
\limsup\sqrt[n]{|a_{n+1}|}=\limsup\sqrt[n]{|a_n|}
\end{equation}
hold?
I' m sure that this is the case, and it may seem a silly question, but it has been bothering me for hours. I have tried using the fact that $$\sqrt[n]{|a_{n+1}|}=\left(\sqrt[n+1]{|a_{n+1}|}\right)^{\frac{n+1}{n}},$$ and other properties of the fractional powers such as $\sqrt[n]{a}<a$ if and only if $a>1$. My main idea was trying to use the squeeze theorem, but I'm stuck.
 A: For each $r \in \mathbb{Z}$, define
$$\alpha(r) = \limsup_{n\to\infty} |a_{n+r}|^{1/n} \in [0, \infty].$$
Let $r, s \in \mathbb{Z}$ be arbitrary. Then we can find a subsequence $(n(k))_{k\geq 1}$ such that $|a_{n(k)+r}|^{1/n(k)}$ tends to $\alpha(r)$. Write $n'(k) = n(k)+r-s$.

*

*Suppose that $\alpha(r) < \infty$. Then
$$|a_{n'(k)+s}|^{1/n'(k)} = \Bigl( |a_{n(k)+r}|^{1/n(k)} \Bigr)^{n(k)/(n(k)+r-s)} \xrightarrow{k\to\infty} \alpha(r)^{1} = \alpha(r), $$
and so, $\alpha(r)$ is a limit point of $|a_{n+s}|^{1/n}$. This proves $\alpha(r) \leq \alpha(s)$.


*Suppose that $\alpha(r) = \infty$. Then we have $|a_{n(k)+r}|^{1/n(k)} \geq 1$ and $\frac{n(k)}{n(k)+r-s} \geq \frac{1}{2}$ for any sufficiently large $k$. So, as $k\to\infty$,
$$ |a_{n'(k)+s}|^{1/n'(k)} = \Bigl( |a_{n(k)+r}|^{1/n(k)} \Bigr)^{n(k)/(n(k)+r-s)} \geq \Bigl( |a_{n(k)+r}|^{1/n(k)} \Bigr)^{1/2} \xrightarrow{k\to\infty} +\infty. $$
This shows that $\alpha(s) = \infty$ as well.
Combining altogether, $\alpha(r) \leq \alpha(s)$ holds unconditionally. However, since $r$ and $s$ are arbitrary, this implies that $\alpha$ is constant. Therefore the desired claim is proved.
A: It suffices to prove that if $(c_{n})$ is a sequence of non-negative numbers such that $\limsup c_{n} = L\in[0,\infty]$ then $\limsup c_{n}^{1+1/n} = L$. First assume $L\in[0, \infty)$ and let $(c_{n_{k}})$ be a subsequence converging to $L$. Then for $L > 0$, $$\lim c_{n_{k}}^{1+1/n_k} = \lim \exp\Big(\Big(1+\frac{1}{n_{k}}\Big)\log(c_{n_{k}})\Big) = \exp\Big(\lim\Big(1+\frac{1}{n_{k}}\Big)\cdot\log(\lim c_{n_{k}})\Big) = L.$$ If $L = 0$, then without loss of generality we may assume that the terms of the subsequence are positive. Since $c_{n_{k}}\rightarrow 0$, $\log(c_{n_{k}})\rightarrow -\infty$ and thus $$\Big(1+\frac{1}{n_{k}}\Big)\log(c_{n_{k}})\rightarrow\ -\infty\implies \exp\Big(\Big(1+\frac{1}{n_{k}}\Big)\log(c_{n_{k}})\Big)\rightarrow 0.$$ Thus, $L$ is the limit of a subsequence of $(c_{n}^{1+1/n})$. Now let $\varepsilon > 0$ and choose $N$ large enough such that $c_{n} < L+\varepsilon$ for all $n\geq N$. Thus, for $n\geq N$ we have $c_{n}^{1+1/n} < (L+\varepsilon)^{1+1/n}$ which implies $$\limsup c_{n}^{1+1/n}\leq L+\varepsilon.$$ Since $\varepsilon$ was chosen to be an arbitrary positive number, we have $\limsup c_{n}^{1+1/n}\leq L$ which shows that every subsquential limit is bounded by $L$ and we're done. Now if $L = \infty$, choose a subsequence such that $c_{n_{k}}\geq k\implies c_{n_{k}}^{1+1/n_k}\geq k^{1+1/n_{k}}\geq k$ which shows that $\limsup c_{n}^{1+1/n} = \infty$.
