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Original question: Provide a sequence ${a_n}$ such that $a_n > 0, \lim\limits_{n→+∞} {a_n}^{1/n}$ exists but $\lim\limits_{n→+∞} \frac{a_{n+1}}{a_n}$ does not exist.

I figured that $\{a_n\}=\sin(n)$ would be a good candidate. Through squeeze theorem, $\lim\limits_{n→+∞} {\sin(n)}^{1/n}=1$ since $\lim\limits_{n→+∞} {-1}^{1/n}=1$ and $\lim\limits_{n→+∞} {1}^{1/n}=1$.

This now leaves me to show that $\lim\limits_{n→+∞} \frac{\sin(n+1)}{\sin(n)}$ does not exist. Using a limit calculator, I found out that this limit is -$\infty$ to +$\infty$. However, I would like to prove this limit does not exist for conciseness.

My work so far: Suppose by contradiction that $\lim\limits_{n→+∞} \frac{\sin(n+1)}{\sin(n)}=L\in\mathbb R$. Then, $\lim\limits_{n→+∞}{ \cos(1) +\cot(n)\sin(1)}=L$. From here, I'm not sure if I can do this or not, $\lim\limits_{n→+∞} \cot(n)=\frac{L-\cos(1)}{\sin(1)}$. I know this would be a contradiction, but I want to delve a bit deeper for a contradiction that is easy to see. Thus, $\lim\limits_{n→+∞} \frac{\cos(n)}{\sin(n)}=\frac{L-\cos(1)}{\sin(1)}$. From here, $\sin(n)$ or $\cos(n)$ must have a limit, which is contradictory. From that we can conclude that the original claim is true: $\lim\limits_{n→+∞} \frac{\sin(n+1)}{\sin(n)}$ does not exist.

Thank you for your time as always.

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    $\begingroup$ You think $a_n=\sin n$ is a good candidate for $a_n>0$ for every $n$? How much is $\sin 5$ already? $\endgroup$ – Did May 7 '17 at 16:41
  • $\begingroup$ Well, I guess I'm back to the drawing board then. Sorry. $\endgroup$ – Arandomuser May 7 '17 at 16:41
  • $\begingroup$ Do you want a correct answer to the original question? $\endgroup$ – Parcly Taxel May 7 '17 at 16:56
  • $\begingroup$ I'll try and find out a solution on my own. Thank you for the offer. $\endgroup$ – Arandomuser May 7 '17 at 16:58
  • $\begingroup$ I found out an actual answer to the original question. I could instead have ${a_n}$ as $|\sin(n)|$. $\endgroup$ – Arandomuser May 7 '17 at 17:49
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A good answer to the original question is $a_n = 2^{\lfloor n/2\rfloor}$. The limit of the quotient of consecutive terms does not exist.

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HINT

$$\sin(n+1) = \sin n \cos 1 + \cos n \sin 1$$ so $$ \frac{\sin(n+1)}{\sin n} = \cos 1 + \sin 1 \cot n $$ and you want to take the limit of the cotangent. What happens to that limit?

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    $\begingroup$ This is precisely the answer I would have provided. $\endgroup$ – Chickenmancer May 7 '17 at 16:43
  • $\begingroup$ Except that, as another answer by the OP themselves, now deleted, shows, they do not know how to prove that, in their words, [as $n$ goes to $+\infty$, $\cot(n)$ goes between $-\infty$ and $+\infty$]... And, if one thinks about it, to give a neat proof of this (true) fact is probably a more involved task than what is needed to answer the question the OP has. To sum it up, if you have a neat argument showing (the weaker result) that the sequence $(\cot n)$ does not converge, you might want to add it to your post. $\endgroup$ – Did May 8 '17 at 5:18

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