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




  • $\begingroup$ The first seems to follow from Bolzano-Weistrass' theorem. $\endgroup$ Nov 8 '14 at 12:14
  • $\begingroup$ @rschwieb I am so sorry,there are two questions. $\endgroup$
    – Eufisky
    Nov 8 '14 at 12:26
  • $\begingroup$ 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? $\endgroup$
    – D. Thomine
    Nov 8 '14 at 12:27
  • $\begingroup$ @Eufisky: I may misunderstand your comment. Do the two limits need to be satisfied for the same subsequence, or not? $\endgroup$
    – D. Thomine
    Nov 8 '14 at 12:31
  • 2
    $\begingroup$ 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 $\lim\limits_{i\to\infty}{\lvert(a_{n_i})^2-a_{n_{i+1}}a_{n_{i-1}}\rvert}^{1/{n_i}}=1$? $\endgroup$ 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.


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