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