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Suppose $a_n\geq 0$, $\varlimsup a_n\leq 1$, $l>1$. then prove $\lim_{n\to \infty}\frac{a_n}{l^n}=0$

Below is my idea.

$\log\frac{a_n}{l^n}=\log a_n-n\log l$, then $$\varlimsup \log\frac{a_n}{l^n}=\varlimsup (\log a_n-n\log l) =\log \varlimsup a_n-\lim n\log l \leq -\infty,$$

so

$$0<\lim \frac{a_n}{l^n}\leq \varlimsup \frac{a_n}{l^n}=e^{\varlimsup \log\frac{a_n}{l^n}}=0,$$

beacuse $\log$ is continuous, we can exchange the order of $\log $ and $\varlimsup$.

Is there a problem with this solution? Thanks for your comments！

Ychen
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    I'm not sure that in general, we can exchange $\limsup$ and $f$ for a continuous function $f$. Instead, you can do the following: since $\ell > 1 = \limsup a_{n}$, pick $N$ large enough such that $a_{n} < \beta < \ell$ for $n\geq N$ where $\beta\in (1, \ell)$. So $a_{n}/\ell < \beta/\ell < 1$ for $n\geq N$ and use the Squeeze Theorem. – Karthik Kannan Dec 09 '23 at 11:59

1 Answers1

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You can use the result lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $

$\displaystyle 0\le\lim\limits_{n\to\infty}\frac{a_n}{l^n}\le\varlimsup\limits_{n\to\infty}\frac{a_n}{l^n}\le\varlimsup\limits_{n\to\infty}a_n\varlimsup\limits_{n\to\infty}\frac{1}{l^n}\le 1\times 0\le 0$

But you can also use If $ \limsup ( a_n ) = L \in \mathbb{R} $, then is $ (a_n) $ bounded from above?

Since $a_n$ bounded and since $\dfrac M{l^n}\to 0$ you have your result.

zwim
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