Bounded sequences

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
We want to prove the intuitively obvious result that if a sequence is increasing and its elements are all smaller than some fixed number, then there is nothing else to happen but for the sequence to have a limit. It has nowhere else to go. We will need to give a name to being smaller than some number (we call this being bounded) and properly define increasing (it’s slightly different to what you might think).

Definition of bounded

The following definitions are key throughout the chapter.

  1. The sequence $(a_n)$ is bounded below if there exists $m\in \R $ such that $m\leq a_n$ for all $n\in \N$. The number $m$ is called a lower bound.
  2. The sequence $(a_n)$ is bounded above if there exists $M\in \R $ such that $a_n\leq M$ for all $n\in \N$. The number $M$ is called an upper bound.
  3. The sequence $(a_n)$ is bounded if $a_n$ is bounded above and below. I.e., there exists $m,M\in \R $ such that $m\leq a_n\leq M$ for all $n\in \N$.


  1. The sequence $n/(n+1)$ is bounded below by $0$, $-1/2$, $-5$ million and bounded above by $1$, $2.3$, $\pi $, and $3$ billion.
    Note that $0$ is the best lower bound in the sense that there’s isn’t a larger one.
    Similarly, $1$ is the best upper bound — there isn’t a smaller one. (The terms in this topic can be confusing. We talk about the smallest largest bound and so on. Pay close to attention to what these phrases mean!)
    Note that we could have said that $n/(n+1)$ is bounded below by $-10$ and bounded above by $3$ as these are equally valid bounds — they are just not the best ones!

  2. The sequence $n$ is bounded below by $0$ but is not bounded above.
  3. The sequence $\cos (n)$ is bounded as $-1\leq \cos (n) \leq 1$. (These bounds are the best possible.)

Our first result on sequences and bounds is that convergent sequences are bounded.
Theorem: Convergent sequences are bounded
If $(a_n)$ is convergent, then $(a_n)$ is bounded above and below.
We shall prove that $(a_n)$ is bounded above. The proof for bounded below is similar.

Suppose the limit of $a_n$ is $a$. By definition we know that given $\varepsilon >0$ there exists $N\in \N$ such that $|a_n-a|<\varepsilon $ for all $n>N$. In particular, taking $\varepsilon =1$ we know that $a_n-a<1$ for all $n>N$. That is, $a_n\lt a+1$ for all $n>N$.
M=\max \{ a_1 , a_2 , a_3 , \dots , a_{N-1}, a_N , a+1 \} .
Then $M$ is such that $a_n\leq M$ for all $n\in \N $. I.e., $M$ is an upper bound.

Write out the proof for bounded below.

A bound on the limit
In this section we show that a bound for a convergent sequence is also a bound for its limit. First, we prove a more general result.
Suppose that $a_n\to a$, $b_n\to b$ and $a_n\leq b_n$ for all $n\in \N$. Then $a\leq b$.
Let $c_n=b_n-a_n$. Then $c_n\geq 0$ for all $n\in \N$ and, by the Algebra of Limits, $c=\lim c_n = \lim b_n – \lim a_n = b-a$.

As $c_n\geq 0$ for all $n$ we have $c\geq 0$, i.e., $b-a\geq 0$.


  1. If $a_n\to a$ and $a_n\leq M$ for all $n\in \N$, then $a\leq M$.
  2. If $a_n\to a$ and $m\leq a_n$ for all $n\in \N$, then $m\leq a$.

(i) Let $b_n=M$.

(ii) This is similar to part (i).

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