Suppose that $(a_n)_n$ and $(b_n)_n$ are two real sequences converging to $a$ and $b$ respectively (both in $\mathbb{R}$). Let $a > b$ and define the sequence $(x_n)_n$ as follows: $$x_{2n} = a_n \quad x_{2n-1} = b_n,$$ that is $(x_n)_n = (a_0, b_1, a_1, b_2, \ldots)$.

I am asked to prove that lim inf of $x_n$ is $b$ and limsup equals $a$.

I first noticed that for some $n_0$ we have that $a_n > b_n$ for all $n \geq n_0$. Indeed, the sequence $(a_n - b_n)_n$ converges to $a-b > 0$ and therefore there is an $n_0 \in \mathbb{N}$ such that $$|a_n - b_n - (a-b)| < \frac{a-b}{2}$$ which is equivalent with $$\frac{a-b}{2} < a_n - b_n < 3\frac{a-b}{2}.$$

In class, we defined lim inf as $\lim_{n \to + \infty} y_n$ with $y_n = \inf\{x_k \vert k \geq n\}$. I think it suffices to look at $\inf\{b_k \vert 2k-1 \geq n\}$ (because $a_n > b_n$ for all $n$) to determine the $y_n$ (I still need to explain why I can discard of the $a_n$ in this infimum) and hence it limit, but I am stuck... The same with limsup. I know both exists, since the sequences $(a_n)_n$ and $(b_n)_n$ are both converging, hence bounded.

Can someone give a hint on this problem? Am I working in the right direction?


2 Answers 2


Let $n_0 \in \mathbb{N}$ be such that $b_n \le a_n, \forall n \ge n_0$.

For $n \ge n_0$ we then have

$$\inf\{x_k : k \ge n\} \ge \inf\{x_{2k-1} : 2k-1 \ge n\} = \inf\{b_k : 2k-1 \ge n\}$$

because $x_{2k-1} \le x_{2k}$.

On the other hand, in general we have

$$\inf\{x_k : k \ge n\} \le \inf\{x_{2k-1} : 2k-1 \ge n\} = \inf\{b_k : 2k-1 \ge n\}$$

because we are taking the infimum over a smaller set.

Therefore $\inf\{x_k : k \ge n\} = \inf\{b_k : 2k-1 \ge n\} = \inf\left\{b_k : k \ge \frac{n+1}{2}\right\}$ for $n \ge n_0$ so

$$\liminf_{n\to\infty} x_n = \lim_{n\to\infty}\inf\{x_k : k \ge n\} = \lim_{n\to\infty}\inf\left\{b_k : k \ge \frac{n+1}{2}\right\} = \liminf_{n\to\infty} b_n = \lim_{n\to\infty} b_n = b$$

The proof of $\limsup_{n\to\infty} x_n = a$ is analogous. Also, you can consider the sequence $(-x_n)_n$ so the above proof shows

$$-a = \liminf_{n\to\infty} -x_n = - \limsup_{n\to\infty} x_n \implies \limsup_{n\to\infty} x_n = a$$

  • $\begingroup$ The equality where you go from lim inf to the regular limit, how can we justify this? $\endgroup$
    – Student
    Jul 15, 2018 at 10:36
  • $\begingroup$ @Student The sequence $(b_n)_n$ is convergent so $\liminf_{n\to\infty} b_n = \limsup_{n\to\infty} b_n = \lim_{n\to\infty} b_n$. $\endgroup$ Jul 15, 2018 at 10:38
  • $\begingroup$ Oh yes, stupid of me. I'll read your answer more carefully when I'm back home tomorrow. Thanks! $\endgroup$
    – Student
    Jul 15, 2018 at 10:39

If $c>a$, then $a_n,b_n<c$ if $n$ is large enough. Therefore, $x_n<c$ if $n$ is large enough and so $\limsup_nx_n\leqslant c$.

On the other hand, if $c<a$, then $a_n>c$ if $n$ is large enough and so $x_n>c$ infinitely often. Therefore, $\limsup_nx_n\geqslant c$.

Since $c>a\implies\limsup_nx_n\leqslant c$ and $c<a\implies\limsup_nx_n\geqslant c$, $\limsup_nx_n=a$.

You can prove that $\liminf_nx_n=b$ in a similar way.

  • $\begingroup$ Thanks for your answer, I'll have to think about this when I'm home again tomorrow. $\endgroup$
    – Student
    Jul 15, 2018 at 10:37

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