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I was presented a problem stating that $n^{10000}/2^n$ converges as $n\to\infty$. I already know that there's a proposition that says the following;

Let $(\epsilon_k)^\infty_{n=0}$ be a null sequence of positive numbers. Then a sequence $(a_n)^\infty_{n=0}$ converges to $l$ if and only if for each $k$ there exists $n_k$ such that $|a_n-l|<\epsilon_k$ for all $n\geq n_k$.

I get kind of the idea of what is being said here, but how do I apply this to sequences in general, and in particular the problem stated above?

PS: is it also true that for any $p>0$ the same holds for $n^p/2^n$?

Marc
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  • Possible duplicate of https://math.stackexchange.com/questions/2489665/prove-that-if-k-in-mathbb-n-and-a1-then-lim-limits-n-to-infty-fra/ – rtybase Mar 27 '18 at 13:12
  • @Marc Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 01 '18 at 07:46
  • @gimusi but it isn't solved. – Marc Apr 02 '18 at 13:54
  • @Marc We simply need to show that $klog x-x \to -\infty$ and we can use l’Hospital. – user Apr 02 '18 at 14:06
  • @gimusi that's simply not my question. – Marc Apr 03 '18 at 18:22

2 Answers2

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Another way is to use $\ln(x)/x \to 0$ as $x \to \infty$ so $\ln(x)-x \to -\infty$.

Then, if $p>0$ and $a>1$, $n^p/a^n =e^{p\ln(n)-n\ln(a)} $. Since $\ln(a)>0$, $p\ln(n)-n\ln(a) \to -\infty$ as $n \to \infty$ so $n^p/a^n \to 0$.

marty cohen
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Note that by ratio test

$$a_n=\frac{n^p}{2^n}\implies \frac{a_{n+1}}{a_n}=\frac{(n+1)^p}{2^{n+1}}\frac{2^n}{n^p}=\frac12\left(\frac{n+1}{n}\right)^p\to \frac12<1\implies a_n\to 0$$

By the definition we need to prove that

$$0<\left|\frac{n^p}{2^n}-0\right|<\epsilon\iff 0<\frac{n^p}{2^n}<\epsilon \iff -\infty<p \log n-n\log 2<\log \epsilon$$

then all reduce to show that $f(x)=k\log x-x\to -\infty$ for $x\to \infty$.

user
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  • @ihf ops...of course, typo fixed thanks! – user Mar 27 '18 at 12:23
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    This is perfect but I fear the OP wants to do this from first principles... – lhf Mar 27 '18 at 12:24
  • Thank you for your answer. This is a clever way of seeing it indeed, however, how can I apply the theorem to conquer the problem? (or is it not practical/possible to do so?)

    Also, what if $p\to\infty$ as well? Then you have $(1+\frac{1}{n})^p$ which converges to $e$, and $\frac{e}{2}\not\lt 1$...

     – Marc Mar 27 '18 at 12:26
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    @Marc note that $p\to\infty$ is not the same as taking $p>0$ arbitrarily large. This works for any fixed $p$, no matter how large it is. –  Mar 27 '18 at 12:28
  • @Marc yes exactly as vrugtehagel explained, p is fixed it doesn't matters how big it is, n is infinitely larger. – user Mar 27 '18 at 12:38
  • $(1+1/n)^p$ converges to 1 by simply using binomial? – Sorfosh Mar 27 '18 at 14:54
  • @Sorfosh simply by continuity $$\left(\frac{n+1}{n}\right)^p\to(1)^p=1$$ – user Mar 27 '18 at 14:56
  • @gimusi but continuity applies to real values. In this case $n \to \infty$ – Sorfosh Mar 27 '18 at 14:58
  • @Sorfosh what is $\frac{n+1}{n}\to?$ for $n\to \infty$? – user Mar 27 '18 at 14:59
  • My goodness I am slow, thank you for your patience. – Sorfosh Mar 27 '18 at 15:03