<s>[INST] Constant sequence limit Let us consider the sequence $(a_n)_{n \ge 1}$ of positive real numbers such that
$$\lim _ {n \to \infty} \left( \frac {a_1}{a_2} + \frac {a_2}{a_3} + \dots + \frac {a_{n-1}}{a_n} + \frac {a_n}{a_1} -n\right)=0.$$
Prove that the sequence is constant.
My try: I considered the $AM -GM$ inequality and I proved that the sequence in parantheses is always greater or equal than $0$. I can't go any further. 
 [/INST] We use induction to prove that $ a_1 = \cdots = a_m$ for all $m = 1, 2, \cdots$. Since the base case is trivial, we focus on the induction step.
Assume that $a_1 = \cdots = a_m$. Then for $n > m$, we have
\begin{align*}
\frac{a_1}{a_2} + \cdots + \frac{a_n}{a_1} - n
&= \overbrace{\left(\frac{a_1}{a_2} + \cdots + \frac{a_{m-1}}{a_m} \right)}^{=m-1} + \left( \frac{a_m}{a_{m+1}} + \cdots + \frac{a_n}{a_1} \right) - n \\
&= \frac{a_m}{a_{m+1}} + \cdots + \frac{a_n}{a_m} - (n-m+1)
\end{align*}
Applying the AM-GM, we find that
\begin{align*}
\frac{a_{m+1}}{a_{m+2}} + \cdots + \frac{a_n}{a_m}
&\geq (n-m)\left( \frac{a_{m+1}}{a_{m+2}} \cdots \frac{a_n}{a_m} \right)^{\frac{1}{n-m}} \\
&= (n-m)\left( \frac{a_{m+1}}{a_m} \right)^{\frac{1}{n-m}}
\end{align*}
So if we let $r = a_m/a_{m+1}$, then it follows from the AM-GM and the assumption that
$$ 0 \leq r + (n-m)r^{-\frac{1}{n-m}} - (n-m+1) \leq \frac{a_1}{a_2} + \cdots + \frac{a_n}{a_1} - n \xrightarrow[n\to\infty]{} 0 $$
and hence by the squeezing lemma,
$$ \lim_{n\to\infty} \left( r + (n-m)r^{-\frac{1}{n-m}} - (n-m+1) \right) = 0. $$
This limit can be explicitly computed, and the result is
$$ \lim_{n\to\infty} \left( r + (n-m)r^{-\frac{1}{n-m}} - (n-m+1) \right) = r - \log r - 1. $$
It is easy to check that the RHS, considered as function of $r$, achieves the unique global minimum $0$ at $r = 1$. This implies $a_{m+1} = a_m$ and hence the induction hypothesis is true for $m+1$ as well.
Therefore the claim follows from the mathematical induction.
</s>