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I am trying to show that the following sequence diverges to $\infty$

$$\left\lbrace \frac{a^n}{n^a} \right\rbrace, \ \; a > 1$$

However, the $a>1$ in combination with a faction is confusing me on how I would show this, could someone demonstrate/explain how to do this.

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Let $u_n=\frac{a^n}{n^a}$ with $a>1$ fixed.

Then :

$$\ln(u_n)=n\ln(a)-a\ln(n)=n\left(\ln(a)-a\frac{\ln(n)}n\right)$$

We know that $\displaystyle{\lim_{n\to\infty}\frac{\ln(n)}n=0}$, hence $\displaystyle{\lim_{n\to\infty}}\left(\ln(a)-a\frac{\ln(n)}n\right)=\ln(a)>0$ and so :

$$\lim_{n\to\infty}\ln(u_n)=+\infty$$

Finally :

$$\lim_{n\to\infty}\frac{a^n}{n^a}=+\infty$$

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Let $c_n:=\frac{n^a}{a^n}$. Then

$c_n^{1/n} \to \frac{1}{a}<1$. Hence the series $\sum c_n$ converges. Therefore $c_n \to 0$.

This gives $\frac{1}{c_n} \to \infty$

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  • $\begingroup$ Solid approach ... (+1) $\endgroup$ – Mark Viola Mar 17 '17 at 19:32

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