If $\lim_n |\alpha_n|=0$ then $\lim_n |\alpha_n|^{1/n}=0$?

Asked May 18 '25 at 03:00

Active May 18 '25 at 03:15

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If $\lim_{n\to\infty} |\alpha_n|=0$ then $\lim_{n\to\infty} |\alpha_n|^{1/n}=0$? I am trying to figure this out.

My attempt, since $\lim_{n\to\infty} |\alpha_n|=0=\lim_{n\to\infty} 1/n^n$, then we have $$ \lim_{n\to\infty} \log|\alpha_n|/n= \lim_{n\to\infty} \log(1/n^n)/n= \lim_{n\to\infty} -n \log(n)/n = -\infty. $$

But I am not sure how we may interchange/switch those limits. As I understand, because $\lim_n |\alpha_n|=0$, the $|\alpha_n|$ must be so close to small numbers like $1/n^n$. Is this limit $\lim_n |\alpha_n|^{1/n}=0$ correct?

edited May 18 '25 at 03:15

Thành Nguyễn

3,512

asked May 18 '25 at 03:00

mathLearner

7

What about $a_n=1/2^n$? – Martin R May 18 '25 at 03:05
2

In general, $\lim_{n\to\infty}|a_n|^{1/n}=0$ implies $\lim_{n\to\infty}|a_n|=0$ but not conversely. Also, why are people downvoting this question? The post owner has shown some honest effort at determining whether this holds or not. – Alann Rosas May 18 '25 at 03:22
$0^0$ is an indeterminate. – Just a user May 18 '25 at 03:32
1

Let $a_n:=-\log|\alpha_n|$. Your question is equivalent to: does $a_n\to\infty$ imply $\frac{a_n}n\to\infty$? The answer should be clear. – Anne Bauval May 18 '25 at 03:50
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Another example: $a_n = 1/n \to 0$, $a_n^{1/n} \to 1$. See https://math.stackexchange.com/q/28348/42969. – Martin R May 18 '25 at 04:34

If $\lim_n |\alpha_n|=0$ then $\lim_n |\alpha_n|^{1/n}=0$?

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