Bounded Sequences

One Bound for Every Term

A sequence is an infinite list of real numbers, so checking that each individual term is finite does not show that the whole sequence stays within a fixed range. Boundedness asks for one number that controls every term, no matter how large the index becomes. This idea will be useful when studying more detailed properties of sequences.

Earlier in the course, a sequence was treated as a function from $\mathbb{N}_0$ to $\mathbb{R}$, and boundedness was related to boundedness of its range. The definition below states the same idea directly in terms of the terms $a_n$.

Definition: A real sequence $(a_n)_{n=0}^{\infty}$ is bounded if there is a real number $M\geq 0$ such that $$ |a_n|\leq M $$ for every $n\in\mathbb{N}_0$. The number $M$ is called a bound for the sequence.

The bound must work for all indices. It is not enough to find a bound for the first several terms, since later terms may be larger. The bound need not be the smallest possible one, either: if $M$ works, then every real number greater than or equal to $M$ also works. A sequence can even have $M=0$ as a bound, which happens precisely when all its terms are zero.

The absolute value in the definition handles positive and negative terms uniformly. Equivalently, $|a_n|\leq M$ means $-M\leq a_n\leq M$. Thus boundedness requires both that the terms do not grow arbitrarily large in the positive direction and that they do not decrease arbitrarily far in the negative direction.

Finding a Bound from a Formula

A useful approach is to manipulate the expression for $a_n$ until it is controlled by a constant independent of $n$. If a formula involves a positive denominator, first check its sign. Then use inequalities that hold for every allowed index, rather than relying only on a few computed terms.

Worked Example: Bounding a Rational Expression

Let $$ a_n=\frac{3n+2}{n+4},\qquad n\in\mathbb{N}_0. $$ Since $n\geq0$, the denominator $n+4$ is positive, and the numerator $3n+2$ is positive. Moreover,

$$ 3(n+4)-(3n+2)=3n+12-3n-2=10>0. $$

Dividing the resulting inequality $3n+2<3(n+4)$ by the positive number $n+4$ gives

0<a_n=\frac{3n+2}{n+4}<3.

Consequently, $|a_n|=a_n<3$ for every $n\in\mathbb{N}_0$, so $M=3$ is a bound. In fact, the inequality is strict, but boundedness only requires $|a_n|\leq M$; it does not require that any term equal the bound.

Worked Example: Bounding an Alternating Sequence

Define $$ b_n=(-1)^n+\frac{2}{n+1},\qquad n\in\mathbb{N}_0. $$ The first term is $b_0=1+2=3$, while $b_1=-1+1=0$. These values alone do not prove boundedness, so we estimate the formula for every index. By the Triangle Inequality,

|b_n| =\left|(-1)^n+\frac{2}{n+1}\right| \leq |(-1)^n|+\left|\frac{2}{n+1}\right|.

For every nonnegative integer $n$, $|(-1)^n|=1$ and $n+1\geq1$, so $0<2/(n+1)\leq2$. Therefore

|b_n|\leq 1+\frac{2}{n+1}\leq 3.

Thus $3$ is a bound for the whole sequence. The estimate does not depend on whether $n$ is even or odd, which is why it covers both signs of the alternating term.

Algebra with Bounded Sequences

Once bounds for individual sequences are known, they can be combined to bound new sequences. For sums, the Triangle Inequality controls the combined terms. For scalar multiples, the absolute value of the scalar determines how much the bound changes.

Theorem: Let $(a_n)$ and $(b_n)$ be bounded real sequences, and let $c\in\mathbb{R}$. Then the sequences $(a_n+b_n)$ and $(ca_n)$ are bounded. More specifically, if $|a_n|\leq M$ and $|b_n|\leq N$ for every $n$, where $M,N\geq0$, then $$ |a_n+b_n|\leq M+N \quad\text{and}\quad |ca_n|\leq |c|M $$ for every $n\in\mathbb{N}_0$.

Proof. Fix any $n\in\mathbb{N}_0$. By the Triangle Inequality and the assumed bounds,

|a_n+b_n|\leq |a_n|+|b_n|\leq M+N.

Since $M+N$ does not depend on $n$, it is a bound for $(a_n+b_n)$. For the scalar multiple, the absolute-value product rule gives

|ca_n|=|c|\,|a_n|\leq |c|M.

Again, $|c|M$ is independent of $n$, so it is a bound for $(ca_n)$. This includes $c=0$ and also allows either original bound to be zero. $\square$

The sum estimate can be applied repeatedly. In particular, any finite linear combination of bounded sequences is bounded: at each step, add the bounds for the sequences and multiply a bound by the absolute value of its scalar coefficient. This gives a systematic way to control expressions built from several sequences.

Worked Example: Combining Two Bounds

Let $$ x_n=(-1)^n+\frac12,\qquad y_n=\frac{1}{n+1}. $$ Because $|(-1)^n|=1$, the Triangle Inequality gives

|x_n| \leq |(-1)^n|+\left|\frac12\right| =1+\frac12 =\frac32.

Also, $n+1\geq1$, so $|y_n|=1/(n+1)\leq1$. The theorem now provides bounds for a sum and a scalar multiple. For example,

|2x_n-y_n| \leq |2x_n|+|-y_n| =2|x_n|+|y_n| \leq 2\left(\frac32\right)+1 =4.

Therefore $(2x_n-y_n)$ is bounded by $4$. The estimate may not be the smallest possible bound, but it is valid for every index and proves the required property.

Products of Bounded Sequences

A product requires a different estimate from a sum. The absolute value of a product is the product of the absolute values, so bounds $M$ and $N$ for the two sequences give the product bound $MN$.

Theorem: If $(a_n)$ and $(b_n)$ are bounded real sequences, then their termwise product $(a_nb_n)$ is bounded. If $|a_n|\leq M$ and $|b_n|\leq N$ for every $n$, with $M,N\geq0$, then $$ |a_nb_n|\leq MN $$ for every $n\in\mathbb{N}_0$.

Proof. Fix $n\in\mathbb{N}_0$. The absolute value product rule and the two assumed bounds give

|a_nb_n|=|a_n|\,|b_n|\leq MN.

The number $MN$ does not depend on $n$, so it is a bound for the product sequence. If either $M=0$ or $N=0$, the same argument gives $|a_nb_n|\leq0$; in that case at least one of the sequences is identically zero, and the product sequence is also identically zero. $\square$

Worked Example: Bounding a Product

Consider $$ p_n=\left((-1)^n+\frac12\right)\frac{1}{n+2}, \qquad n\in\mathbb{N}_0. $$ As in the preceding estimate, $\left|(-1)^n+\frac12\right|\leq3/2$. Since $n+2\geq2$, we also have $0<1/(n+2)\leq1/2$. Hence

|p_n| =\left|(-1)^n+\frac12\right|\frac{1}{n+2} \leq \frac32\cdot\frac12 =\frac34.

Thus $3/4$ is a bound. For a direct check at the first two indices,

p_0=\left(1+\frac12\right)\frac12=\frac34, \qquad p_1=\left(-1+\frac12\right)\frac13=-\frac16.

Both values satisfy the bound, and the product estimate proves that every later term does as well.

When No Single Bound Works

To show that a sequence is not bounded, it is not enough to notice that its terms are increasing. An increasing sequence can remain within a fixed range. Instead, one must show that every proposed bound fails for at least one term. The Archimedean Property, established earlier in the course, is useful for choosing an index large enough to exceed a specified threshold.

Worked Example: An Unbounded Sequence with an Alternating Term

Define $$ c_n=2n+(-1)^n,\qquad n\in\mathbb{N}_0. $$ We show that this sequence is unbounded by ruling out every possible bound. Let $M\geq0$ be any proposed bound. By the Archimedean Property, there is a positive integer $n$ such that $n>(M+1)/2$. Since $(-1)^n\geq-1$, it follows that

c_n=2n+(-1)^n\geq 2n-1>M.

Thus no $M\geq0$ bounds all the terms. In particular, $|c_n|\geq c_n>M$ for this chosen index, so the absolute-value definition of boundedness also fails. The alternating term changes the value by only $1$, while the term $2n$ grows beyond any fixed bound.

What Boundedness Does—and Does Not—Say

Boundedness is a uniform control on size, not a claim that the terms move in one direction or settle down. For example, the sequence $((-1)^n)$ is bounded because $|(-1)^n|=1$ at every index, even though its terms alternate between $1$ and $-1$. Conversely, increasing terms are not automatically bounded; the example $c_n=2n+(-1)^n$ is unbounded.

A bound also need not be a term of the sequence. In the rational-expression example, $3$ is a valid bound even though each term is strictly less than $3$. When proving boundedness, the goal is to produce any fixed bound that works for every index, not to identify an optimal bound or to show that it is attained.

The algebraic results provide reusable tools: bounds for two sequences immediately give a bound for their sum and product, and a scalar multiple has a correspondingly scaled bound. These estimates turn complicated expressions into manageable ones, provided each component has already been controlled uniformly.

Check Your Understanding

Use the definition and the bounds proved in this tutorial to answer the following questions.

For $d_n=(5n+1)/(n+3)$, find a constant that bounds $|d_n|$ for every $n\in\mathbb{N}_0$, and justify the inequality.
Show that $e_n=(-1)^n-3/(n+1)$ is bounded by applying the Triangle Inequality.
If $|u_n|\leq4$ and $|v_n|\leq2$ for every $n$, what bounds follow from this tutorial for $u_n+v_n$, $3u_n$, and $u_nv_n$?
Explain why checking the first five terms of a sequence cannot, by itself, establish that the sequence is bounded.
To show a sequence is unbounded, what must be demonstrated about an arbitrary proposed bound?

What You'll Learn