# Theorems on bounds

Definition.

Let $( a n )$ be a sequence. We say that $( a n )$ is bounded if there are some $M 1 , M 2 ∈ ℝ$ such that $M 1 ⩽ a n ⩽ M 2$ for all $n ∈ ℕ$, i.e., if

This page contains two results on bounds, both very useful. The first says that a convergent sequence is bounded. This is useful because the definition of boundedness is much simpler than the definition of convergence. Convergence has four quantifiers whereas boundedness can be written with only two. It is often easier to show a sequence is not bounded than to show it does not converge to any limit.

Our second result says something about the limit of a convergent sequence when we know bounds for the sequence, but there is a trap. If $A < a n < B$ for all $n$ and $( a n )$ converges to $l$ then the theorem says $A ⩽ l ⩽ B$. (We cannot conclude that $A < l < B$.)

The definition of boundedness of sequences is equivalent to

Definition.

Let $( a n )$ be a sequence. Then $( a n )$ is bounded if there is some $M ∈ ℝ$ such that $| a n | ⩽ M$ for all $n ∈ ℕ$, i.e., if

Theorem on Boundedness of Convergent Sequences.

Let $( a n )$ be a sequence that converges to some $l ∈ ℝ$. Then $( a n )$ is bounded.

Proof.

We assume and must prove .

Remark.

We are given a statement starting and must therefore feed it a positive number. The statement we are trying to prove starts , so we must also define $M$. This proof starts by chosing the easiest possible $ε$ and using this to get $N$ such that all $a n$ are within $1$ of $l$, and using this to give $M$.

Let $ε = 1$.

Let $N ∈ ℕ$ satisfy .

Let $M = max ( | a 0 | , | a 1 | , | a 2 | , … , | a N | , | l | + 1 )$.

Remark.

We now need to show , requiring us to let $n$ be arbitrary.

Subproof.

Let $n ∈ ℕ$ be arbitrary.

If $n < N$ then $| a n | ⩽ M$ since by choice of $M$, $M ⩾ | a n |$.

On the other hand if $n ⩾ N$ then $| a n - l | < ε$, and $| a n | ⩽ | a n - l | + | l |$ by the triangle inequality, so $| a n | < ε + | l | = | l | + 1 ⩽ M$.

Either way we have proved $| a n | ⩽ M$.

Therefore as $n$ was arbitrary.

Therefore .

Example.

For many sequences, such as the ones given by $a n = n$ or $b n = ( -1 ) n ⁢ n$ it is possible to show the sequence is not bounded and therefore the sequence is not convergent. That's a saving of two quantifiers' worth of work. On the other hand, not all non-convergent sequences are unbounded. $c n = ( -1 ) n$ provides an example of a bounded sequence that does not converge to any real number. So this method cannot determine the non-convergence of all sequences and we will require other methods too.

If a sequence is bounded there is the possibility that is has a limit, though this will not always be the case. If it does have a limit, the bound on the sequence also bounds the limit, but there is a catch which you must be careful of.

Theorem giving bounds on limits.

Suppose $( a n )$ is a sequence that converges to some $l ∈ ℝ$. Suppose also that there are $M 1 , M 2 ∈ ℝ$ such that $M 1 < a n < M 2$ for all $n ∈ ℕ$. Then $M 1 ⩽ l ⩽ M 2$.

The catch is that you cannot conclude that $M 1 < l < M 2$. There are a lot of results in analysis like this where a $<$ sign does not carry through to the limit, but becomes $⩽$ instead.

Proof.

We assume that and . We must prove that $M 1 ⩽ l$ and $l ⩽ M 2$.

Remark.

We prove each of these inequalities individually by contradiction.

Subproof.

Assume that $l < M 1$.

Remark.

To use the convergence of $( a n )$ we need to choose a small positive $ε$. Something connected with $l$ and $M 1$ looks appropriate.

Let $ε = M 1 - l$, so $ε > 0$.

Let $N ∈ ℕ$ such that .

Remark.

We try to prove $a n ⩽ M 1$ for some $n$.

Let $n = N$, so since $n ⩾ N$ we have $| a n - l | < ε$.

Now as $a n > M 1 > l$ we have $| a n - l | = a n - l = a n - M 1 + M 1 - l > M 1 - l = ε$ which is a contradiction.

So $M 1 ⩽ l$.

Subproof.

Assume that $l > M 2$.

Let $ε = l - M 2$, so $ε > 0$.

Let $N ∈ ℕ$ such that .

Remark.

Note that this isn't necessarily the same $N$ as before as we have a different $ε$ now.

Let $n = N$, so since $n ⩾ N$ we have $| a n - l | < ε$.

Now as $a n < M 2 < l$ we have $| a n - l | = l - a n = l - M 2 + M 2 - a n > l - M 2 = ε$ which is a contradiction.

So $l ⩽ M 2$.

Example.

Let $a n = 1 - 1 2 n$. So $a n → 1$ as $n → ∞$, as you may verify. (Write down a proof!) Note that $a n < 1$ for all $n$ but the limit is not strictly less than $1$.

Here is another variation of the same result.

Theorem giving bounds on limits.

Suppose $( a n )$ is a sequence that converges to some $l ∈ ℝ$. Suppose also that there are $M 1 , M 2 ∈ ℝ$ such that $M 1 ⩽ a n ⩽ M 2$ for all $n ∈ ℕ$. Then $M 1 ⩽ l ⩽ M 2$.

Proof.

Exercise. Check the proof above works unchanged, in particular that it only needs $M 1 ⩽ a n ⩽ M 2$ for all $n$.