Show that $\lim_{n\to\infty}\frac{\log_an}{n} = 0$ for $0<a<1$

Question

Let $0<a<1$, prove that: $$ \lim_{n\to\infty}\frac{\log_an}{n} = 0 $$

I've started with proving a simpler case for $a>1$. Choose some $\varepsilon >0$ such that: $$ \frac{\log_an}{n} < \varepsilon \iff \log_an < n\varepsilon \iff n<a^{n\varepsilon} $$

We know that for $a>1$ and $k\in \mathbb N$: $$ \lim_{n\to\infty}\frac{n^k}{a^n}=0 $$ Using that fact we may show that: $$ \exists N \in \mathbb N:\forall n \ge N => \frac{n}{(a^\varepsilon)^n} < 1 $$

Since the above yields a true statement starting from $n \ge N$ we may conclude that initial assumption is also true and hence:

$$ \lim_{n\to\infty} \frac{\log_an}{n} = 0 $$

Now consider the case for $0<a<1$: $$ \left|\frac{\log_an}{n}\right| < \varepsilon $$

And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?

Also one of my thoughts was to present $a$ as: $$ a = \frac{1}{1+r} \ \ , r \in \mathbb R $$

and then try to use Bernoulli's, but that didn't yield anything I could use.

Please note that i'm free to use anything before the definition of a derivative.

Update

Based on the hint by MathLover I think this is how I can proceed. Use the fact that:

$$ \log_ax = - \log_{1/a}x $$

Hence: $$ \log_{1/a}n < n\varepsilon $$

Define $b = {1 \over a} > 1$, so:

$$ n < \frac{1}{a^{n\varepsilon}} = b^{n\varepsilon} $$

Now based on the reasoning for case when $a > 1$ we may as well conclude that: $$ \exists N\in \mathbb N : \forall n \ge N \implies \frac{n}{b^{n\varepsilon}} < 1 $$

Thus: $$ \begin{cases} \lim_{n\to\infty}\frac{\log_an}{n} = 0 \\ a\in (0, 1) \bigcup (1, +\infty) \end{cases} $$

Answer 1

We know that for arbitrary function $f$, $$\lim_{n\to\infty}f(n)=0\Longleftrightarrow\lim_{n\to\infty}a^{f(n)}=1$$

So, if $f(n) = \frac{\log_a(n)}n$, then $$\lim_{n\to\infty}\frac{\log_a(n)}n=0\Longleftrightarrow\lim_{n\to\infty}a^{\frac{\log_a(n)}n}=1$$But, if the second limit exists, then $$\lim_{n\to\infty}a^{\frac{\log_a(n)}n}=\lim_{n\to\infty}\bigg(a^{\log_a(n)}\bigg)^{\frac1n}=\lim_{n\to\infty}n^{\frac1n}=1$$

(If you aren't convinced of the last step, see this)

As such, if the limit exists, it has the value of 0.

Answer 2

Hint: $$\log_a(n) = -\log_{1/a}(n).$$

Answer 3

Well, the cases $a<1$ or $a>1$ hardly have any difference as $$\log_an=\frac{\log n}{\log a}$$ and we know that $(1/n)\log n\to 0$.