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Let $a\in [0,1)$. I want to show that $$\lim_{n\to \infty}{na^n}=0$$

My try : $$na^n={n\over e^{-(\log{a})n}}$$ and the limit is $${+\infty\over +\infty}$$ Hence by l'Hopital's rule we have that $$\lim_{n\to \infty}{1\over -(\log{a})e^{-(\log{a})n}}={1\over -\infty}=0$$

Is there any other way to compute this limit ? thanks!

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  • $\begingroup$ Yes! It is a basic limit: exponential functions dominate power functions. $\endgroup$ – Bernard Nov 16 '15 at 12:40
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First, $a$ should be in $(0,1)$, otherwise L'Hopital's Rule won't work until the following step:

$\lim \limits_{n \to \infty} \frac{1}{-(\ln{a})e^{-n\ln{a}}}$

Second, we can use Squeeze Theorem to prove this limit.

Proof :

Let $a = \frac{1}{1+b}$, where $b>0$

$\because na^n > 0$

$na^n=\frac{n}{(1+b)^n}=\frac{n}{1+nb+\frac{n(n-1)}{2}b^2+...}$

$<\frac{n}{\frac{n(n-1)}{2}b^2}=\frac{2}{n-1}$

$\because \lim \limits_{n \to \infty} 0 = 0$

$\lim \limits_{n \to \infty} \frac{2}{n-1} = 0$

By Squeeze Theorem, $\lim \limits_{n \to \infty} na^n =0$

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(Similar) By L'Hopital's Rule $$\lim_{n \to +\infty}na^n=\lim_{n \to +\infty}\frac{n}{\frac{1}{a^n}}\overset{\frac{+\infty}{+\infty}}=\lim_{n \to +\infty}\frac{1}{{-\frac{1}{a^n}\ln(a)}}=-\frac{1}{\ln(a)}\lim_{n \to +\infty} a^n =0$$

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Hint:

Another approach would be to look at the ratio of term $n+1$ to term $n$.

This gives $\frac{(n+1)a^{n+1}}{na^n}=\left(1+\frac{1}{n}\right)a$

Since $a<1$, this factor becomes strictly less than $1$ and then stays bounded away from $1$. Thus eventually the terms of the sequence start decreasing by these factors.

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HINT Expand $e^{-(\log a)n}$ and divide by $n$...

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