For every sequence (z_n) of nonzero complex numbers, show…

For every sequence $(z_n)$ of nonzero complex numbers, show that: $$ \varliminf_{n\to\infty}\left|\frac{z_{n+1}}{z_n}\right|\le\varliminf_{n\to\infty}\sqrt[n]{|z_n|}\le\varlimsup_{n\to\infty}\sqrt[n]{|z_n|}\le\varlimsup_{n\to\infty}\left|\frac{z_{n+1}}{z_n}\right| $$

Step 1 · The right-hand inequality

Let $L=\varlimsup_{n\to\infty}\left|\frac{z_{n+1}}{z_n}\right|$. If $L=\infty$ there is nothing to prove, so assume $L<\infty$. Given $\varepsilon>0$, choose $N$ so that $\left|\frac{z_{k+1}}{z_k}\right|<L+\varepsilon$ for all $k\ge N$. Telescoping for $n>N$:

$$ |z_n|=|z_N|\prod_{k=N}^{n-1}\left|\frac{z_{k+1}}{z_k}\right|<|z_N|(L+\varepsilon)^{n-N} $$

Taking $n$-th roots:

$$ \sqrt[n]{|z_n|}<\sqrt[n]{|z_N|}\cdot(L+\varepsilon)^{1-N/n} $$

As $n\to\infty$: $\sqrt[n]{|z_N|}\to1$ (fixed $N$) and $(L+\varepsilon)^{1-N/n}\to L+\varepsilon$. Hence $\varlimsup_n\sqrt[n]{|z_n|}\le L+\varepsilon$ for every $\varepsilon>0$, so:

$$ \varlimsup_{n\to\infty}\sqrt[n]{|z_n|}\le L=\varlimsup_{n\to\infty}\left|\frac{z_{n+1}}{z_n}\right| $$

Step 2 · The left-hand inequality

Let $\ell=\varliminf_{n\to\infty}\left|\frac{z_{n+1}}{z_n}\right|$. By the symmetric argument (choosing $N$ so $\left|\frac{z_{k+1}}{z_k}\right|>\ell-\varepsilon$ for $k\ge N$, and using $\ell-\varepsilon>0$ when needed):

$$ |z_n|>|z_N|(\ell-\varepsilon)^{n-N} \ \Longrightarrow\ \sqrt[n]{|z_n|}>\sqrt[n]{|z_N|}(\ell-\varepsilon)^{1-N/n}\ \xrightarrow[n\to\infty]{}\ \ell-\varepsilon $$

so $\varliminf_n\sqrt[n]{|z_n|}\ge\ell-\varepsilon$ for every $\varepsilon>0$, giving:

$$ \varliminf_{n\to\infty}\left|\frac{z_{n+1}}{z_n}\right|\le\varliminf_{n\to\infty}\sqrt[n]{|z_n|} $$

Step 3 · The middle inequality

Trivial from the definitions: for any real sequence $(a_n)$, $\ \inf_{k\ge n}a_k\le\sup_{k\ge n}a_k$ for every $n$, so taking $n\to\infty$ preserves $\varliminf a_n\le\varlimsup a_n$.

Example showing the inequalities can be strict

Let $z_n=2^n$ for even $n$, $z_n=3^n$ for odd $n$. Then $\left|\frac{z_{n+1}}{z_n}\right|$ alternates between $\frac32\left(\frac32\right)^n\to\infty$ (odd-to-even transitions) and $2\left(\frac23\right)^n\to0$ (even-to-odd transitions), so:

$$ \varliminf\left|\frac{z_{n+1}}{z_n}\right|=0, \qquad \varlimsup\left|\frac{z_{n+1}}{z_n}\right|=\infty $$

while $\sqrt[n]{|z_n|}$ is exactly $2$ (even $n$) or $3$ (odd $n$), giving $\varliminf\sqrt[n]{|z_n|}=2,\ \varlimsup\sqrt[n]{|z_n|}=3$ — genuinely strict inequalities all the way across the chain: $0<2\le3<\infty$.