# Does $x_{n, m_n} \overset{n \to \infty}{\to} x$ hold, if $m_n = o(n)$ and $x_{n,m} \overset{n \to \infty}{\to} x_m \overset{m \to \infty}{\to} x$?

Suppose we have $$(x_{n,m}) \subset \mathbb R$$ and $$(x_m)\subset \mathbb R$$ and $$x \in \mathbb R$$ such that

• For all $$m \in \mathbb N$$ we have $$x_{n,m} \overset{n \to \infty}{\longrightarrow} x_m \,$$; and
• $$x_m \overset{m \to \infty}{\longrightarrow} x$$.

Let $$(m_n)_n \subseteq \mathbb N$$ be an non-decreasing sequence of indices such that $$m_n \overset{n \to \infty}{\longrightarrow} \infty$$ and $$m_n = o(n)$$.

Does it then hold that $$x_{n, m_n} \overset{n \to \infty}{\longrightarrow} x$$?

My intuition (not rigorous):

I think this could be correct, since $$m_n = o(n)$$ means that $$m_n$$ is kind of "standing still" compared to $$n$$, so $$\lim_{n \to \infty} x_{n, m_n} = \lim_{m \to \infty} \lim_{n \to \infty} x_{n, m}$$.

Thoughts/Attempt:

Using the triangle inequality, we get $$|x_{n,m_n} - x| \leq |x_{n, m_n} - x_{m_n}| + |x_{m_n} - x|.$$ We know that $$|x_{m_n} - x| \overset{n \to \infty}{\longrightarrow} 0$$. Therefore, it would suffice to show $$|x_{n, m_n} - x_{m_n}| \overset{n \to \infty}{\longrightarrow} 0$$.

I'm hoping that $$m_n = o(n)$$ will guarantee than this converges to zero. But I don't know if that's the case.

Possible problem: What if $$(|x_{n,m} - x_m|)_n$$ converges increasingly slowly as $$m$$ gets larger? Could the $$|x_{n,m_n} - x_{m_n}|$$ then not converge to zero?

Note: For my purposes at the moment, I just need this for numbers in $$[0,1]$$. So I could replace $$\mathbb R$$ above with $$[0,1]$$. (Maybe this works, but only on compact sets?)

The little-o assumption doesn't change much. The convergence can be even slower than the growth of $$m_n$$ (whatever $$m_n$$ is!).

To see this, first consider the sequence $$y_{n,m} = \begin{cases} 0 & \text{if } n \le m, \\ 1 & \text{if } n > m. \end{cases}$$ Clearly, $$y_{n,m} \xrightarrow{n \to \infty} 1$$ for each $$m$$, but the subsequence $$y_{n,n}$$ is constantly $$0$$ and doesn't converge to $$1$$. This corresponds to $$m_n = n$$, which is excluded in your question.

So let $$m_n$$ be any non-decreasing sequence such that $$m_n \xrightarrow{n \to \infty} \infty$$ and $$\frac{m_n}{n} \xrightarrow{n \to \infty} 0$$ (although the latter won't play any role). We will construct a counterexample $$x_{n,m}$$ as follows. First, define the sequence $$t_m := \max \{ n : m_n \le m \}.$$ One should check from the definition that

• for each $$m$$, $$\{ n : m_n \le m \}$$ is a bounded subset $$\mathbb{N}$$ and thus has a maximum,
• $$t_{m_n} \ge n$$ for each $$n$$,
• $$t_m \xrightarrow{m \to \infty} \infty$$.

Now we define: $$x_{n,m} := y_{n,t_m}.$$ Finally, it's easy to convince yourself that $$x_{n,m_n} = y_{n,t_{m_n}} = 0 \not\to 1,$$ thanks to $$n \le t_{m_n}$$.

The construction is probably clearer if we distinguish between $$n$$ and $$m$$. Let's use the notation $$y^{(m)} = (y_{1,m},y_{2,m},y_{3,m},\ldots),$$ so that we have a sequence $$y^{(m)}$$ (and each term is a sequence itself). Then the sequence $$x^{(m)}$$ is chosen from $$y^{(m)}$$ as a subsequence: $$x^{(m)} = y^{(t_m)}$$. It's not exactly a subsequence, as $$t_m$$ doesn't have to be strictly increasing, but in principle $$t_m$$ grows fast enough to compensate for the slowness of $$m$$. This last fact is encapsulated in the condition $$t_{m_n} \ge n$$.

• Thanks! This more than answers the question. But for rigor, don't we need to define $t_m$ with the supremum? After all, $t_{m_n} = \max \mathbb R$ isn't well defined, but $\sup \mathbb R = \infty$. Commented Sep 17, 2021 at 8:23
• In the case you described, supremum wouldn't help, as $t_n = \infty$ doesn't make any sense. The definition is OK the way it is, but you're right that it requires justification. Note that $m_n \to \infty$ and hence $\{ n : m_n \le m \}$ is a bounded subset of $\mathbb{N}$. This shows that it indeed has a maximum. Commented Sep 17, 2021 at 10:36

Let $$x_{n,m}:=\left(1-\frac1m\right)^{\log n}.$$ Then $$x_{n,m}\xrightarrow[n\to\infty]{}0=:x_m$$, and $$x_m\xrightarrow[m\to\infty]{}0=:x.$$ However, for $$m_n:=\lceil\sqrt n\rceil=o(n)$$, $$x_{n,m_n}=\exp\left(\log n\cdot\log\left(1-\frac1{\lceil\sqrt n\rceil}\right)\!\right)=\exp\left(-\frac{\log n}{\sqrt n}+o\!\left(\frac{\log n}{\sqrt n}\right)\!\right)\xrightarrow[n\to\infty]{}1.$$

• Hey @nejimban! Thanks for the counterexample. Would you mind explaining why $\log n \cdot \log \left(1 - \frac{1}{\lceil \sqrt n \rceil} \right) = - \frac{\log n}{\sqrt n} + o\left( \frac{\log n}{ \sqrt n} \right)$? Commented Sep 17, 2021 at 8:15
• @zxmkn $\log(1-x)=-x+o(x)$ as $x\to0$. Commented Sep 17, 2021 at 8:24