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Prove that if $k \in \mathbb {N}$ and $a>1$, then $\lim\limits_{n\to\infty} \frac{n^k}{a^n}=0$

I have this, I want to know if what I did is correct.

Let be $a^n>1>0$, then:

$1>\frac{1}{a^n}$

Then:

$|\frac{1}{a^n}|<1$

Aplying archimedean property,for $ε>0$ exists $n \in \mathbb {N}$ such that:

$nε>n^{k+1}$

Then:

$ε>\frac{n^{k+1}}{n}=n^k$

Let be $ε>0$, exists $N=max(1,ε)$ that if $n>N$

$|\frac{n^k}{a^n}-0|=|\frac{n^k}{a^n}|$

$=\frac{n^k}{a^n}$

$<ε (1)= ε$

Therefore the sequences converges to zero.

Thank, you.

J. Jenkins
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  • The archimedean property argument is incorrect, for example, let $\epsilon = \frac{1}{2}$. – Mr. T Oct 25 '17 at 19:57
  • Unfortunately, not: Your proof is not correct. Archimedean property does not imply that there exists $n$ such that $n \epsilon > n^{k + 1}$, since both sides depend on $n$. After all, this would imply $\epsilon > n^k \ge 1$, which is a contradiction whenever $\epsilon < 1$. –  Oct 25 '17 at 19:59

2 Answers2

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You must prove that for any $\varepsilon >0$ there exists $N\in\mathbb{N}$ such that for $n>N$ it happens that $$\frac{n^k}{a^n}<\varepsilon$$ This is quite hard to prove directly. To prove that $$\lim_{n\to\infty} \frac{n^k}{a^n}=0$$ We can use L'Hopital rule $k$ times and find that $$\lim_{n\to\infty} \frac{n^k}{a^n}=\lim_{n\to\infty} \frac{kn^{k-1}}{a^n(\log a) }=\ldots=\lim_{n\to\infty} \frac{k!}{a^n(\log a)^k}=0$$

Hope this helps

Raffaele
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Let's assume for a second the following is true $$\lim\limits_{n\rightarrow \infty}\frac{\ln{n}}{n}=0 \tag{1}$$

Now, let's look at the original problem from a different perspective, e.g. $$\ln{\left(\frac{n^k}{a^n}\right)}=\ln{\left(n^k\right)}-\ln{\left(a^n\right)}=k\ln{\left(n\right)}-n\ln{\left(a\right)}$$ $$k\ln{\left(n\right)}-n\ln{\left(a\right)}<-k\ln{\left(n\right)} \Leftrightarrow 2k\ln{\left(n\right)} < n\ln{\left(a\right)} \Leftrightarrow \frac{\ln{n}}{n}<\frac{\ln{a}}{2k} \tag{2}$$ given $(1)$ is true, $(2)$ is true as well from some $n$ onwards. But then, from some $n$ onwards we have $$\ln{\left(\frac{n^k}{a^n}\right)} < -k\ln{\left(n\right)} \Leftrightarrow 0<\frac{n^k}{a^n} < \frac{1}{e^{k\ln{n}}}=\frac{1}{n^k}$$ and the result follows from the squeeze theorem.

How do we prove $(1)$ though? Well, there are plenty of examples of this proof, for example this one.

rtybase
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