Upper and Lower Bounds for Sequences

Bounds in Each Direction

The previous tutorial defined a bounded sequence by requiring one fixed number to control the absolute value of every term. That definition combines two distinct requirements: the terms must not rise arbitrarily high, and they must not fall arbitrarily low. It is often useful to separate these requirements. An upper bound controls the first direction, while a lower bound controls the second.

A bound is a number that works for every index, not just for a finite collection of terms. It also need not be one of the terms of the sequence. The definitions make these points explicit.

Definition: Let $(a_n)_{n=0}^{\infty}$ be a real sequence. A real number $U$ is an upper bound for the sequence if $$ a_n\leq U $$ for every $n\in\mathbb{N}_0$. The sequence is bounded above if it has an upper bound. A real number $L$ is a lower bound for the sequence if $$ L\leq a_n $$ for every $n\in\mathbb{N}_0$. The sequence is bounded below if it has a lower bound.

The direction of the inequality matters: every term must lie at or below an upper bound, and at or above a lower bound. A sequence can have many upper bounds and many lower bounds. If $U$ is an upper bound, then any $V\geq U$ is also an upper bound, since $a_n\leq U\leq V$ for every $n$. Likewise, if $L$ is a lower bound, any $K\leq L$ is also a lower bound.

Two One-Sided Bounds Give Boundedness

Having an upper bound alone does not prevent terms from becoming arbitrarily negative. Having a lower bound alone does not prevent them from becoming arbitrarily positive. Having both bounds places every term inside one fixed interval.

Theorem: A real sequence is bounded if and only if it is both bounded above and bounded below.

Proof. First suppose $(a_n)$ is bounded. By the definition of boundedness, there is an $M\geq0$ such that $|a_n|\leq M$ for every $n$. The absolute-value inequality implies

-M\leq a_n\leq M

for every $n$. Thus $-M$ is a lower bound and $M$ is an upper bound.

Conversely, suppose the sequence has a lower bound $L$ and an upper bound $U$. Then $L\leq a_n\leq U$ for every $n$. Let $M=\max\{|L|,|U|\}$, which is nonnegative. From $a_n\leq U\leq |U|\leq M$ and $L\leq a_n$, we also have $-M\leq L\leq a_n$, because $-M\leq -|L|\leq L$. Therefore

-M\leq a_n\leq M

for every $n$, so $|a_n|\leq M$. The sequence is bounded. $\square$

This theorem lets us choose the most convenient form of a proof. To establish boundedness, we may find one absolute-value bound as in the previous tutorial, or we may find a lower and an upper bound separately. The second approach can be easier when the formula naturally gives an interval containing all the terms.

Worked Example: Bounds for a Rational Sequence

For $n\in\mathbb{N}_0$, define $$ a_n=\frac{4n-1}{n+2}. $$ Since $n+2>0$, we can rewrite the expression as

a_n=\frac{4(n+2)-9}{n+2} =4-\frac{9}{n+2}.

Because $n+2\geq2$, we have $0<9/(n+2)\leq9/2$. Subtracting these inequalities from $4$ gives

-\frac12=4-\frac92\leq a_n<4.

Thus $-1/2$ is a lower bound and $4$ is an upper bound. The lower bound is attained: $a_0=(4(0)-1)/(0+2)=-1/2$. The upper bound is not attained, because $9/(n+2)>0$ for every $n$, so $a_n<4$ for every index. The theorem shows that the sequence is bounded; in particular, the displayed interval also gives $|a_n|\leq4$.

Transforming Bounds

Once the terms are confined to an interval, simple algebra can give bounds for a shifted or rescaled sequence. Adding the same number to every term shifts both endpoints by that number. Multiplying by a positive number preserves the order. Multiplying by a negative number reverses the order, so the endpoints exchange roles.

Theorem: Suppose $L\leq a_n\leq U$ for every $n\in\mathbb{N}_0$, and let $c,d\in\mathbb{R}$. If $c>0$, then $$ d+cL\leq d+ca_n\leq d+cU. $$ If $c<0$, then $$ d+cU\leq d+ca_n\leq d+cL. $$ If $c=0$, then $d+ca_n=d$ for every $n$.

Proof. Fix any $n\in\mathbb{N}_0$. If $c>0$, multiplying $L\leq a_n\leq U$ by $c$ preserves both inequalities, giving $cL\leq ca_n\leq cU$. Adding $d$ throughout gives $d+cL\leq d+ca_n\leq d+cU$.

If $c<0$, multiplication by $c$ reverses both inequalities, giving $cU\leq ca_n\leq cL$. Adding $d$ gives $d+cU\leq d+ca_n\leq d+cL$. Finally, if $c=0$, then $d+ca_n=d+0a_n=d$ for every $n$. Each conclusion holds for every index, so the stated endpoints are bounds. $\square$

The separate cases prevent a common error: multiplying an inequality by a negative number does not preserve its direction. The result also explains how to get bounds without returning to the original formula for every term.

Worked Example: A Negative Scalar Reverses the Bounds

Suppose a sequence satisfies $2\leq a_n\leq5$ for every $n$, and define $b_n=3-2a_n$. Here $c=-2<0$ and $d=3$, so the negative-scalar case gives

3+(-2)(5)\leq b_n\leq3+(-2)(2),

-7\leq b_n\leq-1.

For instance, if $a_n=2$, then $b_n=3-2(2)=-1$, the upper endpoint. If $a_n=5$, then $b_n=3-2(5)=-7$, the lower endpoint. These substitutions confirm why the endpoints switch when the multiplier is negative.

Worked Example: Separate Bounds for an Alternating Sequence

Let $$ b_n=\frac{(-1)^n}{n+1},\qquad n\in\mathbb{N}_0. $$ If $n$ is even, then $b_n=1/(n+1)$, so $0<b_n\leq1$, since $n+1\geq1$. If $n$ is odd, then $n\geq1$, and $n+1\geq2$. Consequently,

-\frac12\leq-\frac{1}{n+1}=b_n<0.

The even and odd cases together show that $-1/2\leq b_n\leq1$ for every $n$. Both bounds are attained: $b_0=1$ and $b_1=-1/2$. Therefore the sequence is bounded. Checking both parity cases is necessary here because the sign of the term changes with the index.

One-Sided Boundedness and Unboundedness

A sequence may have a bound in one direction but not the other. For example, the nonnegative integers viewed as a sequence are bounded below by $0$, but they have no upper bound. To prove that no upper bound exists, the argument must address an arbitrary proposed upper bound and find a term that exceeds it.

Worked Example: A Lower Bound Without an Upper Bound

Consider $c_n=n$ for $n\in\mathbb{N}_0$. Since $n\geq0$, the number $0$ is a lower bound. To show there is no upper bound, let $U\in\mathbb{R}$ be any proposed upper bound. By the Archimedean Property, there is a positive integer $n$ such that $n>U$. For this index,

$$ c_n=n>U. $$

Thus $U$ does not bound the sequence above. Since this applies to every real $U$, the sequence is not bounded above. It is therefore not bounded, despite having a lower bound.

For a sequence to be unbounded, it is enough for one side to fail: it may have no upper bound, no lower bound, or neither. In contrast, boundedness requires both directions to be controlled. A useful proof strategy is to state which direction is in question, take an arbitrary proposed bound in that direction, and then choose an index whose term violates it.

Bounds Need Not Be Attained

An upper or lower bound is not automatically the largest or smallest term of a sequence. It only has to lie on the appropriate side of every term. As the first worked example showed, an upper bound can be strictly greater than every term. Similarly, a lower bound can be strictly less than every term.

When a bound is attained, some term equals it. When it is not attained, every term satisfies a strict inequality at that endpoint. These are different facts from the existence of a bound itself. For example, proving $a_n<4$ for every $n$ establishes that $4$ is an upper bound, but it does not show that any term equals $4$. Nor does it show, by itself, that $4$ is the least possible upper bound.

The key question in a bounds argument is always whether one fixed number works for every index. A handful of computed terms may suggest suitable endpoints, but only an inequality valid for all $n$ proves that those endpoints are bounds. The definitions also keep one-sided claims distinct: an upper bound says nothing on its own about how negative the terms might become, and a lower bound says nothing on its own about how large they might become.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What does it mean for a real number $U$ to be an upper bound for a sequence? How does the definition of a lower bound differ?
If $L\leq a_n\leq U$ for every $n$, explain why $(a_n)$ is bounded in the absolute-value definition.
A sequence satisfies $-3\leq a_n\leq2$ for every $n$. Give lower and upper bounds for the sequence $4-a_n$, and identify which original endpoint gives each new endpoint.
Why does multiplying an inequality by a negative number require reversing its direction?
How would you prove that a sequence has no upper bound, rather than merely checking that its first several terms grow?