# A difficult contest question from the former Soviet Union

Let $(a_n)$ be a positive sequence such that $\varlimsup\limits_{n\to\infty} a_n^{1/n}=1$ and $\varliminf\limits_{n\to\infty} a_n^{1/n}<1$.

Prove there exists a subsequence $(a_{n_i})$ such that

$$\lim\limits_{i\to\infty}\left(a_{n_i}\right)^{1/{n_i}}=1$$

and

$$\lim\limits_{i\to\infty}{\lvert(a_{n_i})^2-a_{n_i+1}a_{n_i-1}\rvert}^{1/{n_i}}=1.$$

• The first seems to follow from Bolzano-Weistrass' theorem. Nov 8 '14 at 12:14
• @rschwieb I am so sorry,there are two questions. Nov 8 '14 at 12:26
• Note that, using the first, the second boils down to $\lim_{i \to +\infty} \left| 1- \frac{a_{n_i+1}a_{n_i-1}}{a_{n_i}^2}\right|^\frac{1}{n_i} = 1$, so $\left|\frac{a_{n_i+1}a_{n_i-1}}{a_{n_i}^2} \right| \ll \frac{1}{n_i}$, or $\left|\frac{a_{n_i}^2}{a_{n_i+1}a_{n_i-1}} \right| \gg n_i$. It might be easier to use $\left| 2 \ln (a_{n_i}) - \ln (a_{n_i+1}) - \ln (a_{n_i-1}) \right| \gg \ln(n_i)$, which looks like a bound on a second derivative. Assume that $\left| 2 \ln (a_{n_i}) - \ln (a_{n_i+1}) - \ln (a_{n_i-1}) \right| \leq C \ln(n_i)$, and find a contradiction? Nov 8 '14 at 12:27
• @Eufisky: I may misunderstand your comment. Do the two limits need to be satisfied for the same subsequence, or not? Nov 8 '14 at 12:31
• Is that really $\lim\limits_{i\to\infty}{\lvert(a_{n_i})^2-a_{n_i+1}a_{n_i-1}\rvert}^{1/{n_i}}=1$ and not $\lim\limits_{i\to\infty}{\lvert(a_{n_i})^2-a_{n_{i+1}}a_{n_{i-1}}\rvert}^{1/{n_i}}=1$? Feb 20 '19 at 4:24

Define $$A= \{(k_n)_{n\ge 1} ; \lim_{n \rightarrow \infty }{k_n^{\frac{1}{n}}}=1 \land ( \forall i\ge 1 ; k_i \in (a_n)_{n \ge 1} ) \}$$ , and consider the sequence below: $$\bigcup_{(d_n)_{n \ge 1} \in A} (d_n)_{n \ge 1} = (c_{i_j})_{i_j \ge 1}$$ now we prove this sequence can be accepted as an answer , first one can observer that $$(c_{i_j})_{i_j \ge 1} \subsetneq (a_n)_{n \ge 1}$$ since $$(a_n)_{n \ge 1}$$ does not converge to 1 , and $$(c_{i_j})_{i_j \ge 1}$$ is countable since it is subsequence of $$(a_n)_{n \ge 1}$$.
now consider sequences which can be created in the form of $$(c_{i_j-1})_{i_j-1 \ge 1}$$ and $$(c_{i_j+1})_{i_j+1 \ge 1}$$ we proof that neither of this sequences does not converge to 1.which mean after finitely steps they do not converge to $$1$$ anymore.
assume by contrary one of them for instance $$(c_{i_j-1})_{i_j-1 \ge 1}$$ converges to $$1$$ this means $$(c_{i_j-1})_{i_j-1 \ge 1} \in \bigcup_{(a_n)_{n \ge 1} \in A} (a_n)_{n \ge 1}$$ now again we can repeat the same argue for the two obtained subsequences $$(c_{i_j-2})_{i_j-2 \ge 1}$$ and $$(c_{i_j})_{i_j \ge 1}$$ , one can easily see if this go infinitely that mean the sequence $$(a_n)_{n \ge 1}$$ converges to $$1$$ which is a contrary to the problem assumption.
now set $$M = \sup|c_{i_j}^2 - c_{i_j-1}c_{i_j+1}|$$ and $$m = \inf|c_{i_j}^2 - c_{i_j-1}c_{i_j+1}|$$ since all of terms are positive and neither of subsequences $$(c_{i_j})_{i_j-1 \ge 1}$$ and $$(c_{i_j})_{i_j+1 \ge 1}$$ does not converges to $$1$$ so their multiplication does not too , we can conclude that there exist index like $$j_0$$ such for any $$j_0 \le i_j$$ we have $$0 < m \le 1$$ and $$0 < M \le 1$$ now we can write : $$\limsup_{j \rightarrow \infty} |c_{i_j}^2-c_{i_j-1}c_{i_j+1}|^{\frac{1}{i_j}} \le \lim_{j \rightarrow \infty } M^{\frac{1}{i_j}} = 1$$ and $$\liminf_{j \rightarrow \infty} |c_{i_j}^2-c_{i_j-1}c_{i_j+1}|^{\frac{1}{i_j}} \ge \lim_{j \rightarrow \infty } m^{\frac{1}{i_j}} = 1$$ which force: $$\liminf_{j \rightarrow \infty} |c_{i_j}^2-c_{i_j-1}c_{i_j+1}|^{\frac{1}{i_j}}=\limsup_{j \rightarrow \infty} |c_{i_j}^2-c_{i_j-1}c_{i_j+1}|^{\frac{1}{i_j}}=1$$ This subsequence also has the first property which ends the proof.