More on Sequences — Question 2

*Question 2 Let $(b_n)$ be a bounded real sequence and define $a_n=\sup\{b_k:k\ge n\}$.

1. Explain why every $a_n$ exists.

2. Prove that $(a_n)$ is decreasing and bounded, and hence convergent.

3. State what its limit represents, and compute it when $b_n=(-1)^n+1/n$.

*Question 2 – Solution

Let $T_n=\{b_k:k\ge n\}$. Each tail is nonempty, and because $(b_n)$ is bounded, $T_n$ is bounded above. The least-upper-bound property of $\mathbb R$ therefore guarantees that $a_n=\sup T_n$ exists.

Removing the first element of a tail cannot increase its set of possible values. Formally, $T_{n+1}\subseteq T_n$, so $$a_{n+1}=\sup T_{n+1}\le\sup T_n=a_n.$$ Thus $(a_n)$ is decreasing.

If $m$ is any lower bound for $(b_n)$, then $m\le b_k\le a_n$ for every $k\ge n$, so $m\le a_n$. Hence $(a_n)$ is bounded below. The Monotone Convergence Theorem now gives convergence, and its limit is called the limit superior: $$\lim_{n\to\infty}a_n=\limsup_{n\to\infty}b_n.$$

For $b_n=(-1)^n+1/n$, even terms equal $1+1/(2j)$ and odd terms are negative. Among the even terms in a tail, the earliest one is largest. If $e_n$ is the first even integer satisfying $e_n\ge n$, then $$a_n=1+\frac1{e_n}\longrightarrow1.$$ Thus $\limsup b_n=1$. The tail-supremum process has turned oscillating bounded data into a decreasing upper envelope.