# Is there a sequence $x_n\to+\infty$ such that $\liminf x_{2n}/x_n = 0$?

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?

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.

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

• Isn't it should be $(x_{2^kn_0})_k$ ? Jul 16, 2019 at 14:56
• This answers the original question as it appeared in the title (but not the body). The question has since been corrected. Jul 16, 2019 at 14:58
• okay, since I bugged you I deserve some up votes on my question. Come on guys... xD Jul 16, 2019 at 15:06
• @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. Jul 16, 2019 at 15:08
• 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. Jul 16, 2019 at 15:22