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Consider a sequence with strictly positive terms $(a_n)_{n\geq1}$ with the property: $$\lim_{n\rightarrow \infty} \left(\frac{a_1}{a_2}+\frac{a_2}{a_3} + \cdots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}-n\right)=0$$

Prove that this sequence is constant.

It's a problem that I have no idea. Maybe someone has an idea? Thank you!

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  • $\begingroup$ I'm starting by supposing it's constant except for one term: $$a,a,a,\ldots, a, A, a,a,\ldots, a,\ldots$$ Then we have a sum $$ 1+1+1+\cdots + 1 + \frac a A + \frac A a + 1 + \cdots + 1 - (\text{number of foregoing terms}). $$ $$ = \frac a A + \frac A a - 2 > 0 \text{ unless } A=a. $$ $\endgroup$ – Michael Hardy Dec 15 '16 at 20:41
  • $\begingroup$ BTW, to show that $$\frac a A + \frac A a > 2 \text{ unless }A=a,$$ "complete the square": $$ \frac a A + \frac A a = \left(\frac a A - 2 + \frac A a \right) + 2 = \left( \sqrt{\frac a A} - \sqrt{\frac A a} \, \right)^2 + 2. $$ $\endgroup$ – Michael Hardy Dec 15 '16 at 20:43
  • $\begingroup$ Next suppose it's constant except for three consecutive terms:$$a, a, a, \ldots, a, a, A, B, a, a, a, \ldots, a, a$$ Then we'd want to show $$\frac a A + \frac A B + \frac B a > 3 \text{ unless } a= A = B.$$ $\endgroup$ – Michael Hardy Dec 15 '16 at 20:48
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    $\begingroup$ Any finite number of terms is easy by AM-GM. the difficulty is in proving the sequence is eventually constant $\endgroup$ – cats Dec 15 '16 at 20:49
  • $\begingroup$ So the point would be to try to show that all deviations from constancy make the sum preceding the $\text{“} \cdots-n \text{''}$ bigger. $\qquad$ $\endgroup$ – Michael Hardy Dec 15 '16 at 20:50
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Suppose the sequence is not constant and suppose WLOG $\frac{a_2}{a_1} = 1 + \epsilon > 1$ (just pick the first term that is less than the next, which has to exist as otherwise everything is at least $1$ and the difference is bounded away from $0$). Now note by AM-GM that $$\require{align} \require{cancel} \begin{align} \frac{a_1}{a_2} + \left(\frac{a_2}{a_3} + \ldots + \frac{a_n}{a_1}\right) - n &= \frac{1}{1+\epsilon} + \left(\frac{a_2}{a_3} + \ldots + \frac{a_n}{a_1}\right) - 1 - (n-1) \\ &\ge \left(\frac{1}{1+\epsilon}-1\right) + (n-1)\sqrt[n-1]{\frac{a_2}{\cancel{a_3}}\frac{\cancel{a_3}}{\cancel{a_4}}\cdots\frac{\cancel{a_n}}{a_1}} - (n-1) \\ &= \frac{1}{1+\epsilon} - 1 + (n-1)\cdot\left( \sqrt[n-1]{1+\epsilon}-1\right). \end{align}$$

In the limit, the right term approaches $\ln(1+\epsilon)$ while the left is fixed, so this difference is bounded away from $0.$

Just for completeness, I'll give a way of seeing that this is actually positive. First note that each expression is positive by weighted AM-GM. Therefore in the limit we get something nonnegative. Now evidently $\ln(1+\epsilon) \not= 1-\frac{1}{1+\epsilon},$ so it's positive.

Addendum: the limit is $\ln(1+\epsilon)$ by a standard application of l'Hopital's, as $$\lim_{x\to\infty} \frac{a^{1/x} - 1}{1/x} = \lim_{x\to\infty} \ln a \cdot \frac{\cancel{\frac{-1}{x^2}}a^{1/x}}{\cancel{\frac{-1}{x^2}}} = \ln a.$$

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  • $\begingroup$ split $n$ into $n-1+1$ and AM-GM is just on the $(n-1)$ terms in parentheses $\endgroup$ – cats Dec 16 '16 at 3:52

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