Lim Sup and Lim Inf

This is a question with a few components to it. In each one, I give my attempt at a solution. Thank you in advance for any suggestions/answers/advice on how to solve this problem.

For bounded sequences which do not converge, these are notions which sometimes can substitute for the non-existent limit.

Let $\{a_n\}$ be a bounded sequence. Define $T_n = \{a_i : i \geq n\}$ denote the n-th "tail" of the sequence and let $\overline{b}_n = \sup T_n$ and $\underline{b}_n = \inf T_n$.

a). Prove the sequences $\{\overline{b}_n\}$ and $\{\underline{b}_n\}$ both converge.

Given that $\{a_n\}$ is a bounded sequence and $\overline{b}_n = \sup T_n$ and $\underline{b}_n = \inf T_n$, then we have for the supremum, by defintion, that there exists a least upper bound, call it $\overline{m}$, such that $x \leq \overline{m}$ for all $x \in \{a_n\}$. Similarly, there exists a greatest lower bound $\underline{m}$ such that $x \geq \underline{m}$ for all $x \in \{a_n\}$. This is where I am stuck because I want to say that because $\overline{b}_n$ is given to be our $\sup T_n$, then we have a monotone sequence which converges by the Completeness Property. But is this a safe assumption at this point? This doubt is the same for the latter case, except we change our argument so that it states it using $\underline{b}_n$ as the $\inf T_n$.

b). We now define $\lim \sup a_n = \lim \overline{b}_n$, and $\lim \inf a_n = \lim \underline{b}_n$. Find $\lim \sup$ and $\lim \inf$ for the sequence $a_n = \frac{1}{n} + (-1)^n$.

For this part of the problem, I have an intuitive understanding about the $\lim \sup a_n = 1,$ and the $\lim \inf a_n = -1$, but am not sure how to formally state this.

c). Prove that $\lim \inf a_n \leq \lim \sup a_n$.

Similar to above, intuitively, I believe that this is true since the $\inf a_n$ is the greatest lower bound and that $\sup a_n$ would be the least upper bound. However, I then consider the empty set as a counterexample (is this valid)?

d). Prove that: $\lim a_n$ exists $\iff$ $\lim \inf a_n = \lim \sup a_n$.

As far as this step, I would not know how to get an 'if and only if' proof for $\lim \inf a_n = \lim \sup a_n$. The only thing that I could think of is knowing from previous courses that the limit is unique, and that it must approach the same value from either side.

Again, thank you very much for taking the time to read/answer/post on this rather lengthy question. I am using the textbook Introduction to Analysis by Arthur Mattuck.

1 Answer

Hint for (a): both sequences are monotone and bounded.

Hint for (b): you can explicitely find $\bar b_n$ and $\underline b_n$. Your intuition is correct, btw.

Hint for (c): $\bar b_n\ge\underline b_n$ always.

Hint for (d): $\bar b_n\ge a_n\ge\underline b_n$ to prove in one direction and use the fact that a convergent sequence satisfies the Cauchy criterion to prove in another direction.

Anothe way to understand $\limsup$ and $\liminf$ is to take all convergent subsequences; then find $\sup$ and $\inf$ of the set of their limits. This approach, for example, instantly gives the answer to (b).

• Thank you very much for this @TZakrevskiy. For part d), isn't the Cauchy Criterion something along the lines of assuming without loss of generality that if $n>m$, then $| a_n - a_m |$? Or is there still something missing here? Commented Sep 25, 2013 at 2:26
• @Jamil_V Cauchy criterion says "$\{a_n\}$ converges $\iff$ $\forall \varepsilon>0\,\exists N>0: \forall n,m>N$ we have $|a_n-a_m|<\varepsilon$" Commented Sep 25, 2013 at 2:28
• On your comment about proving d), we can use $\bar b_n\ge a_n\ge\underline b_n$ because $\{b_n\}$ is a subsequence right? Commented Sep 25, 2013 at 3:36
• @Jamil_V No, the sequence $\{\bar b_n\}$ is not a subsequence of $\{a_n\}$ (same for $\{\underline b_n\}$), you can build a counterexample. We have those two inequalities because $\bar b_n = \sup \{a_i:\,i\ge n\}\ge a_n\ge \inf \{a_i:\,i\ge n\}= \underline b_n$. Commented Sep 25, 2013 at 10:31