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?
 A: 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 $(x_{p(n)})_n$.
A: 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.
