Upper Bounds, Lower Bounds, And Boundedness

Why Bounds Matter

A sequence can be studied by asking how its terms are ordered, and also by asking how far those terms can extend. The second question is about bounds. A bound is a number that constrains every term of a sequence, not merely the first few terms.

Bounds are useful because they give a precise way to describe the range of a sequence even when its terms do not move consistently upward or downward. They also provide one of the key hypotheses in the Monotone Convergence Theorem. Here, the goal is to define bounds accurately and learn to establish them directly from inequalities.

Upper Bounds and Lower Bounds

Let \((a_n)\) be a real sequence, with terms \(a_n\) indexed by \(n\in\mathbb{N}\). An upper bound must be at least as large as every term; a lower bound must be at most as large as every term.

Definition: A real number \(U\) is an upper bound for \((a_n)\) if \(a_n\le U\) for every \(n\in\mathbb{N}\). A real number \(L\) is a lower bound for \((a_n)\) if \(L\le a_n\) for every \(n\in\mathbb{N}\). The sequence is bounded above if it has an upper bound, and bounded below if it has a lower bound.

A sequence is bounded if it is both bounded above and bounded below. Equivalently, there are real numbers \(L\) and \(U\) such that every term lies in the interval from \(L\) to \(U\):

L\le a_n\le U \qquad\text{for every }n\in\mathbb{N}.

The inequalities in this definition need not be strict. A bound may equal a term, or it may be larger or smaller than every term. Also, bounds are not usually unique: if \(U\) is an upper bound, then any larger real number is an upper bound too. The definition concerns all terms, including terms far beyond those we have calculated.

Absolute-Value Form of Boundedness

It is often convenient to express boundedness using absolute value. The following equivalence explains precisely when that form can be used.

Theorem: A real sequence \((a_n)\) is bounded if and only if there is a real number \(C\ge 0\) such that \(|a_n|\le C\) for every \(n\in\mathbb{N}\).

Proof. First suppose \(|a_n|\le C\) for every \(n\), where \(C\ge0\). The absolute-value inequality implies \(-C\le a_n\le C\). Thus \(-C\) is a lower bound and \(C\) is an upper bound, so the sequence is bounded.

Conversely, suppose the sequence is bounded. Then there are real numbers \(L\) and \(U\) such that \(L\le a_n\le U\) for every \(n\). Choose \(C=\max\{|L|,|U|\}\), which is nonnegative. Since \(L\le a_n\le U\), the value of \(a_n\) lies between numbers whose absolute values are at most \(C\). Therefore \(-C\le a_n\le C\), and hence \(|a_n|\le C\) for every \(n\). This proves both directions.

This result is especially useful when an expression for \(|a_n|\) is simpler to estimate than the sequence itself. The constant \(C\) need not be the smallest possible one; it only needs to work for every index.

Finding Bounds by Comparing Expressions

For an explicit sequence, a standard approach is to compare its formula with a proposed bound. The crucial requirement is that the comparison hold for every allowed index. Algebraic rearrangement often reduces the task to checking that a simple expression is nonnegative.

Worked Example: Bounding a Rational Sequence

Let \(a_n=\dfrac{3n+2}{n+1}\), for \(n\in\mathbb{N}\). Find a lower bound and an upper bound.

For the lower bound, compare \(a_n\) with \(2\). Because \(n+1>0\), we can multiply the comparison by \(n+1\) without changing its direction:

\frac{3n+2}{n+1}\ge 2 \quad\Longleftrightarrow\quad 3n+2\ge 2(n+1) \quad\Longleftrightarrow\quad n\ge 0.

Since \(n\in\mathbb{N}\), the final inequality holds. Thus \(2\le a_n\).

For the upper bound, again use \(n+1>0\):

\frac{3n+2}{n+1}\le 3 \quad\Longleftrightarrow\quad 3n+2\le 3(n+1) \quad\Longleftrightarrow\quad 2\le 3.

The last statement is true, so \(a_n\le3\). Therefore \(2\le a_n\le3\) for every positive integer \(n\). As a check, at \(n=1\), \(a_1=(3+2)/(1+1)=5/2\), and indeed \(2\le5/2\le3\). The inequalities above establish the bounds for all terms, not just this first one.

The sign of a quantity matters when multiplying inequalities. In this example, \(n+1\) is positive, so the inequality direction is preserved. Multiplication by a negative number would reverse the direction. Checking this condition is part of a valid bounds argument.

Boundedness Is Not Monotonicity

Monotonicity describes how consecutive terms are ordered: a sequence is nondecreasing if \(a_n\le a_{n+1}\) for every \(n\), and nonincreasing if \(a_{n+1}\le a_n\) for every \(n\). Boundedness instead asks whether all the terms fit between two fixed real numbers. Neither property implies the other.

Worked Example: Bounded but Not Monotone

Consider \(a_n=(-1)^n\dfrac{n}{n+1}\), for \(n\in\mathbb{N}\). Since \(n>0\) and \(n+1>n\), we have \(0<\dfrac{n}{n+1}<1\). Also, \(|(-1)^n|=1\), so

|a_n| = \left|(-1)^n\right|\frac{n}{n+1} = \frac{n}{n+1} <1.

It follows that \(-1<a_n<1\) for every \(n\), so the sequence is bounded. To check that it is not monotone in either direction, calculate its first terms:

a_1=-\frac12,\qquad a_2=\frac23,\qquad a_3=-\frac34.

The first step is upward because \(-1/2<2/3\), while the next is downward because \(2/3>-3/4\). Thus the sequence is neither nondecreasing nor nonincreasing. The bound is verified by the inequality for every \(n\), while the change in direction is already visible in these terms.

Worked Example: Monotone but Not Bounded

Let \(b_n=n\), for \(n\in\mathbb{N}\). Since \(b_{n+1}=n+1>n=b_n\), the sequence is strictly increasing. However, it is not bounded above. To see this, take any proposed real upper bound \(U\). There is a natural number \(n>U\), so \(b_n=n>U\). Thus \(U\) fails to be an upper bound. Since this argument applies to every proposed \(U\), no upper bound exists.

The first few terms are \(1,2,3,4\), but these values do not establish an upper bound. For instance, \(4\) is not an upper bound because \(b_5=5>4\). The argument for arbitrary \(U\) is what establishes unboundedness, rather than a finite list of terms.

These examples show why the words must be kept distinct. A bounded sequence can move up and down, and an increasing sequence can continue beyond every fixed upper bound. Monotonicity tells us about direction; boundedness tells us about the extent of the terms.

Using an Algebraic Inequality

A useful technique is to transform the desired comparison until it becomes a familiar nonnegative expression, such as a square. This can prove a bound even when the formula does not immediately show one.

Worked Example: A Bound from a Perfect Square

For \(c_n=\dfrac{n}{n^2+1}\), find bounds valid for every \(n\in\mathbb{N}\).

First, \(n>0\) and \(n^2+1>0\), so \(c_n>0\). For an upper bound, test \(1/2\). The denominator is positive, so

\frac{n}{n^2+1}\le\frac12 \quad\Longleftrightarrow\quad 2n\le n^2+1 \quad\Longleftrightarrow\quad 0\le n^2-2n+1 \quad\Longleftrightarrow\quad 0\le(n-1)^2.

A square is nonnegative for every real \(n\), so the comparison holds for every positive integer \(n\). Therefore \(0<c_n\le1/2\), giving a lower bound of \(0\) and an upper bound of \(1/2\). For a direct check at \(n=1\), \(c_1=1/(1+1)=1/2\), which satisfies both inequalities with equality at the upper bound. The square argument proves the upper bound for all indices.

A Common Mistake: Checking Only Initial Terms

A frequent error is to calculate several terms, notice that they lie below a proposed number, and conclude that the number is an upper bound. But an upper bound must control every term. For example, the calculation \(1,2,3<4\) for the first three terms of \(b_n=n\) does not show that \(4\) is an upper bound. In fact, \(b_5=5>4\).

The correction is to return to the definition. To prove that \(U\) is an upper bound, establish \(a_n\le U\) for an arbitrary \(n\in\mathbb{N}\). To show that a sequence is not bounded above, show that every proposed \(U\) is exceeded by some term. Initial terms can help reveal a pattern or expose a false claim, but they cannot certify a bound for all indices.

Key takeaway: A bound is a uniform inequality that holds for every term. To establish it, use an argument valid for arbitrary \(n\); to establish unboundedness, show that any proposed bound fails.

Check Your Understanding

Use the definitions and inequalities from this tutorial to answer the following questions.

What two inequalities must hold for every \(n\) in order to show that \(L\) is a lower bound and \(U\) is an upper bound?
Explain why a sequence satisfying \(|a_n|\le C\) for every \(n\), with \(C\ge0\), is bounded.
For \(d_n=\dfrac{2n+1}{n+1}\), verify whether \(1\) is a lower bound and \(2\) is an upper bound.
Give a reason that boundedness does not imply monotonicity, using the alternating sequence in this tutorial.
What must be shown to prove that a sequence has no upper bound?