Constant sequence? 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!
 A: 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 $$\lim_{x\to\infty} \frac{a^{1/x} - 1}{1/x} = \lim_{x\to\infty} \ln a \cdot \frac{\cancel{\frac{-1}{x^2}}a^{1/x}}{\cancel{\frac{-1}{x^2}}} = \ln a.$$
