Real Analysis: Sequences

Limit Supremum and Limit Infimum

Limit Supremum and Limit Infiumum

Consider a sequence $\{a_n\}$. The limit supremum of $\{a_n\}$ is defined as $\limsup\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} (\sup\limits_{m \geq n} \{a_{m}\})$. Plainly, the supremum asks for the least upper bound of the subsequence caused by removing the first $n$ elements from the original sequence, and then the limit asks what value this sequence of least upper bounds approaches. The limit infimum of $\{a_n\}$ is analogously defined as $\liminf\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty}(\inf\limits_{m \geq n} \{a_{m}\}).$

The limit supremum is also called the limit superior or the upper limit, and the limit infimum is also called the limit inferior or the lower limit. We will stick to limit supremum and limit infimum here.

Doubled Bounded Limit Theorem

The double bounded limit theorem states that the limit of a sequence exists if and only if its limit supremum and limit infimum both exist and are equal to one another. This theorem should not be confused with the theorem that states that a sequence has a limits if and only if all of its subsequential limits are equal; while similar, one deals with limits of subsequences, while the other deals with limits of infima and suprema.

Problems

  1. Let $\{a_n\} = \left\{ \begin{array}{ll} 1 & x \text{ is even}\\\frac{3}{x} & x \text{ is odd}\end{array} \right.$

    1. Graph the first several terms of $\{a_n\}.$

    2. Determine $\limsup\limits_{n \rightarrow \infty}\{a_n\}$ and $\liminf\limits_{n \rightarrow \infty}\{a_n\}.$ Are they equal?

    1. Graph of first few terms of a_n

    2. Note that for all $n > 3,$ $a_n \leq 1$ by definition, and so $\limsup\limits_{n \rightarrow \infty}\{a_n\} = 1.$
      To determine the limit inferior, we partition the original sequence into its even and odd-numbered subsequences. The odd-numbered subsequence is always $1$ by definition, and so its infimum is $1.$ The even-numbered subsequence is $\frac{3}{n}.$ We can see that $\inf\left\{\frac{3}{n}\right\} = 0.$ From here we note that $\inf\{0, 1\} = 0,$ and so $\liminf\{a_n\} = 0.$

    Show Answer

  2. Consider two sequences $\{a_n\}$ and $\{b_n\}$ where $a_n \leq b_n.$ Show the following:

    1. $\liminf\limits_{n \rightarrow \infty}\{a_n\} \leq \liminf\limits_{n \rightarrow \infty} \{b_n\}.$

    2. $\limsup\limits_{n \rightarrow \infty}\{a_n\} \leq \limsup\limits_{n \rightarrow \infty} \{b_n\}.$

    Consider $N \in \mathbb{N}.$ For all $n > N,$ it follows by construction that $s_k \leq t_k.$ Therefore $\inf\limits_{k>N}{s_k} \leq \inf\limits_{k>N}{t_k}$ and $\sup\limits_{k>N}{s_k} \leq \sup\limits_{k>N}{t_k}.$ By the algebraic limit theorem, it follows that $\liminf\limits_{n \rightarrow \infty}\{a_n\} \leq \liminf\limits_{n \rightarrow \infty}\{b_n\}$ and $\limsup\limits_{n \rightarrow \infty}\{a_n\} \leq \limsup\limits_{n \rightarrow \infty}\{b_n\}.$

    Show Answer

  3. Double Bounded Limit Theorem: Show that $\{a_n\}$ converges to $p$ if and only if the limit supremum and limit infimum of $\{a_n\}$ both exist and $\limsup\limits_{n \rightarrow \infty} a_n = \liminf\limits_{n \rightarrow \infty} a_n.$

    Assume $\{a_n\}$ converges to $p$. Then for every $\varepsilon > 0$, there exists and $N \in \mathbb{N}$ such that $d(a_n, p) < \varepsilon$ for every $n > N.$ Therefore $p - \varepsilon < a_n < p + \varepsilon.$ Therefore $\sup \{a_{n>N}\} \leq p + \varepsilon$ and $\inf\{a_{n > N}\} \geq p - \varepsilon$. It follows that $|\sup \{a_{n>N}\} - p| \leq \varepsilon$ and $|\inf\{a_{n > N} - p\} \leq \varepsilon$. Therefore $\limsup\limits_{n \rightarrow \infty} \{a_n\} = p = \liminf\limits_{n \rightarrow \infty} \{a_n\}.$

    Conversely, assume $\limsup\limits_{n \rightarrow \infty} \{a_n\} = p = \liminf\limits_{n \rightarrow \infty} \{a_n\}.$ Then for any $\varepsilon > 0$, there exist some $N, M \in \mathbb{N}$ such that $|\sup\{a_n\} - p| < \varepsilon$ and $|\inf\{a_m\} - p| < \varepsilon$ for all $n > N$ and $m > M$. Pick $K = \text{max}(N, M)$ and such that $|\sup\{a_k\} - p| < \varepsilon$ and $|\inf\{a_k\} - p| < \varepsilon$ whenever $k > K.$ It follows that $p - \varepsilon < a_k < \sup\{a_k\} < p + \varepsilon$ and $p - \varepsilon < \inf\{a_k\} < a_k < p + \varepsilon.$ This simplifies to $p - \varepsilon < a_k < p + \varepsilon,$ which by definition means that $\lim\limits_{k \rightarrow \infty} \{a_k\} = p.$

    Show Answer