I have a sequence of real numbers $(x_n)$ that diverges to $+\infty$. Can I conclude somehow that $$\liminf \frac{x_{2n}}{x_n}>0,$$ or are there counterexamples?


2 Answers 2


Hint: If

$$ x_n = \begin{cases}n, & \text{if $n$ is even,} \\ n^2, & \text{if $n$ is uneven,} \end{cases}$$

what's the limit of

$$\frac{x_{2(2n+1)}}{x_{2n+1}}\;\; ?$$

However, it is indeed impossible that $\frac{x_{2n}}{x_n}$ itself converges to zero. The proof given by mechanodroid is entirely correct. For the problem with taking subsequences, see my comment below mechanodroid's answer.

  • $\begingroup$ That's not true, half of the $x_{2n}/x_n$ will be 2. $\endgroup$ Jul 16, 2019 at 14:47
  • $\begingroup$ @VasilyMitch Yeah it is true , take $n_k=2k+1$ $\endgroup$
    – Filburt
    Jul 16, 2019 at 14:48
  • $\begingroup$ @VasilyMitch That doesn't matter, the smallest limit point is still 0. $\endgroup$ Jul 16, 2019 at 14:49
  • $\begingroup$ TS was asking about the limit first. Your example doesn't have one. Which doesn't matter now. $\endgroup$ Jul 16, 2019 at 14:52
  • $\begingroup$ ok my bad, corrected... $\endgroup$
    – Filburt
    Jul 16, 2019 at 14:54

Answer to the original question whether it is possible that $x_n \to +\infty$ but $\frac{x_{2n}}{x_n} \to 0$.

Assume that $\frac{x_{2n}}{x_n} \to 0$. Then there exists $n_0 \in \mathbb{N}$ such that $n \ge n_0 \implies \left|\frac{x_{2n}}{x_n}\right| \le 1$ or $|x_{2n}| \le |x_n|$.

Therefore $$|x_{n_0}| \ge |x_{2n_0}| \ge |x_{4n_0}| \ge \cdots$$

so $(x_{2^kn_0})_k$ is a (by absolute value) decreasing subsequence of $(x_n)_n$ so $x_n \not\to +\infty$.

Tentative answer to whether it is possible that $x_n \to +\infty$ but $\liminf_{n\to\infty}\frac{x_{2n}}{x_n} \to 0$

Assume that $\liminf_{n\to\infty} \frac{x_{2n}}{x_n} \to 0$. Then there exists a subsequence $(x_{p(n)})_n$ such that $\frac{x_{2p(n)}}{x_{p(n)}} \to 0$ so by the previous part there is a further subsequence of $(x_n)_n$ which is decreasing so $x_n \not\to +\infty$.

However, this is wrong because the constructed decreasing subsequence is not necessarily a subsequence of $(x_{p(n)})_n$.

  • $\begingroup$ Isn't it should be $(x_{2^kn_0})_k$ ? $\endgroup$
    – Bumblebee
    Jul 16, 2019 at 14:56
  • 1
    $\begingroup$ This answers the original question as it appeared in the title (but not the body). The question has since been corrected. $\endgroup$
    – TonyK
    Jul 16, 2019 at 14:58
  • 1
    $\begingroup$ okay, since I bugged you I deserve some up votes on my question. Come on guys... xD $\endgroup$
    – Filburt
    Jul 16, 2019 at 15:06
  • 1
    $\begingroup$ @Filburt I agree, but I'm out of upvotes for today. I'll be able to upvote again in about $8$ hours so remind me then haha. $\endgroup$ Jul 16, 2019 at 15:08
  • 2
    $\begingroup$ The proof that $\frac{x_{2n}}{x_n}$ cannot converge to zero is correct. The problem appears when going over to a subsequence. Note that $|x_{2\rho(n)}|\le |x_{\rho(n)}|$ does not help in constructing a non-increasing subsequence of $(x_{\rho(n)})_n$, as $\rho(n)$ might always be uneven and hence $2\rho(n)\neq \rho(m)$ for all $m,n\in\Bbb N$. It is for essentially the same reason, that my example below cannot be modified in such a way that it contradicts your first statement. So, to sum things up: Yes, 0 can be a limit point. No, 0 cannot be the limit. $\endgroup$ Jul 16, 2019 at 15:22

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