Constant Sequences

When Every Term Has the Same Value

The examples in the previous tutorial showed several ways to specify a sequence: by an explicit formula, by a recurrence, or by a repeating pattern. A particularly simple sequence keeps the same value at every index. This case is worth identifying precisely: a constant sequence is not merely a sequence whose first few terms agree, but one for which the defining rule gives the same value at every nonnegative integer index.

Definition: A sequence $(a_n)_{n=0}^{\infty}$ is constant if there is a real number $c$ such that $a_n=c$ for every $n\in\mathbb{N}_0$. The number $c$ is the constant value of the sequence.

For instance, the formula $a_n=4$ defines a constant sequence with value $4$, while $b_n=(-1)^n$ does not: its terms alternate between $1$ and $-1$. A constant sequence has a range consisting of exactly one real number, namely its constant value. Conversely, if the range contains exactly one real number, every term must equal that number, so the sequence is constant.

The value of a constant sequence is unique. Indeed, if $a_n=c$ and $a_n=d$ for every $n$, then at index $0$ we have $c=a_0=d$. Thus there is no ambiguity in referring to the constant value.

Worked Example: Recognizing a Constant Formula

Define $u_n=\frac{3n+6}{3}$ for $n\in\mathbb{N}_0$. Simplifying the formula gives

u_n=\frac{3n+6}{3}=n+2.

This sequence is not constant: for example, $u_0=2$ and $u_1=3$, so its values at those two indices differ. A formula can simplify substantially without becoming constant; the index still appears in the simplified expression.

In contrast, define $v_n=\frac{2n+8}{n+4}$ for $n\in\mathbb{N}_0$. Since $n+4\geq4>0$, the denominator is never zero, and

v_n=\frac{2(n+4)}{n+4}=2

for every allowed index $n$. Hence $(v_n)$ is constant with value $2$. The nonzero-denominator check matters: the simplification is valid here because $n+4$ cannot vanish for $n\in\mathbb{N}_0$.

Equal Neighbors Determine the Whole Sequence

For an explicit formula, constancy may follow from simplification. For a sequence described by a recurrence, it can be more direct to compare neighboring terms. If each term equals the next one, then every term equals the initial term. This conclusion relies on the rule holding at every index, not just on checking a finite list of neighbors.

Theorem: A real sequence $(a_n)_{n=0}^{\infty}$ is constant if and only if $a_{n+1}=a_n$ for every $n\in\mathbb{N}_0$.

Proof. First suppose that the sequence is constant with value $c$. Then $a_n=c$ and $a_{n+1}=c$ for every $n$, so $a_{n+1}=a_n$.

For the converse, suppose $a_{n+1}=a_n$ for every $n\in\mathbb{N}_0$. We prove by induction that $a_n=a_0$ for every $n\in\mathbb{N}_0$. At $n=0$, the statement $a_0=a_0$ holds. Now let $n\in\mathbb{N}_0$ and suppose $a_n=a_0$. The assumed equality of neighboring terms gives

a_{n+1}=a_n=a_0.

Thus the statement holds at $n+1$. By the Principle of Mathematical Induction, $a_n=a_0$ for every nonnegative integer $n$. The sequence is constant with value $a_0$. $\square$

The theorem provides a useful way to prove constancy without finding a separate formula for every term. It also explains why checking a few adjacent equalities is not enough: the induction step needs the equality $a_{n+1}=a_n$ to hold for every index.

Worked Example: A Constant Sequence Given Recursively

Let $x_0=-\sqrt{5}$, and define $x_{n+1}=x_n$ for every $n\in\mathbb{N}_0$. The recurrence says exactly that neighboring terms are equal. By the theorem, $(x_n)$ is constant. Its value is $x_0=-\sqrt{5}$, so

x_0=-\sqrt{5},\qquad x_1=-\sqrt{5},\qquad x_2=-\sqrt{5},\qquad x_n=-\sqrt{5}\quad\text{for every }n\in\mathbb{N}_0.

The first three terms illustrate the rule, but the theorem justifies the conclusion for every index. In particular, no additional assumption about the sign or size of the starting value is needed.

Constant Sequences and Sequence Operations

Constant sequences remain simple under the basic operations performed term by term. If one sequence always has value $A$ and another always has value $B$, then their sum always has value $A+B$, their difference always has value $A-B$, and their product always has value $AB$. Multiplying every term of a constant sequence by a fixed real number also gives a constant sequence.

Theorem: Let $a_n=A$ and $b_n=B$ for every $n\in\mathbb{N}_0$, where $A,B\in\mathbb{R}$. For every $n\in\mathbb{N}_0$, the sequences defined term by term satisfy $$ a_n+b_n=A+B,\qquad a_n-b_n=A-B,\qquad a_nb_n=AB. $$ If $\lambda\in\mathbb{R}$, then $\lambda a_n=\lambda A$ for every $n$, and $|a_n|=|A|$ for every $n$.

Proof. Fix any $n\in\mathbb{N}_0$. By the definitions of the two constant sequences, $a_n=A$ and $b_n=B$. Substitution gives

a_n+b_n=A+B,\qquad a_n-b_n=A-B,\qquad a_nb_n=AB.

The same substitution gives $\lambda a_n=\lambda A$ and $|a_n|=|A|$. Since $n$ was arbitrary, all of these equalities hold at every index. Each resulting sequence is therefore constant, with the value indicated by its formula. $\square$

Division also preserves constancy when the divisor is a nonzero constant. If $B\neq0$, then $a_n/b_n=A/B$ at every index. The nonzero condition is essential: division by zero is undefined, so it does not produce a real sequence.

Worked Example: Combining Constant Sequences

Let $a_n=7$ and $b_n=-2$ for every $n\in\mathbb{N}_0$. Their sum, difference, product, and a scalar multiple can be calculated at any index using the constant values:

a_n+b_n=7+(-2)=5,\qquad a_n-b_n=7-(-2)=9, $$ $$ a_nb_n=7(-2)=-14,\qquad 3a_n=3(7)=21.

For example, at $n=0$, $a_0+b_0=7+(-2)=5$; at $n=4$, $a_4+b_4=7+(-2)=5$ as well. The theorem guarantees the same calculation at every index, so each of the resulting sequences is constant.

When a Recursive Rule Produces a Constant Sequence

A recurrence can depend on the current term without leaving that term unchanged. To decide whether a constant sequence can satisfy such a rule, substitute the proposed constant value into the recurrence. The value must reproduce itself under the rule. This is a fixed-value condition, and it gives both a necessary and a sufficient test when the starting value is specified.

Theorem: Let $r,c\in\mathbb{R}$, and let $(x_n)$ satisfy $x_{n+1}=r x_n+c$ for every $n\in\mathbb{N}_0$. The sequence is constant if and only if its starting value satisfies $x_0=r x_0+c$.

Proof. Suppose first that $(x_n)$ is constant. Then $x_1=x_0$. Applying the recurrence at $n=0$ gives $x_1=r x_0+c$, so $x_0=r x_0+c$.

Conversely, suppose $x_0=r x_0+c$. We prove by induction that $x_n=x_0$ for every $n\in\mathbb{N}_0$. This holds at $n=0$. If $x_n=x_0$, then the recurrence and the assumed equation give

x_{n+1}=r x_n+c=r x_0+c=x_0.

Thus $x_{n+1}=x_0$, completing the induction step. The sequence is constant with value $x_0$. $\square$

Worked Example: Checking a Constant Solution to a Recurrence

Consider the recurrence $y_{n+1}=\frac{1}{3}y_n+4$, with $y_0=6$. The starting value satisfies the fixed-value condition:

\frac{1}{3}y_0+4=\frac{1}{3}(6)+4=2+4=6=y_0.

The theorem therefore shows that $y_n=6$ for every $n\in\mathbb{N}_0$. Direct checks at the first two steps agree:

y_1=\frac{1}{3}(6)+4=6,\qquad y_2=\frac{1}{3}(6)+4=6.

The induction-based conclusion is stronger than these checks: it establishes the same value at every later index. If the initial value had not satisfied $x_0=r x_0+c$, the theorem would rule out constancy for that particular recursively defined sequence.

What Constancy Does—and Does Not—Say

A constant sequence has one value in its range, and equal neighboring terms provide a complete test for constancy. These descriptions are equivalent, but they emphasize different ways of working: the range description is useful when a formula has been simplified, while the neighboring-term test is often effective for a recurrence. For a recurrence of the form $x_{n+1}=r x_n+c$, the fixed-value equation gives a direct test using the initial term.

Constancy should not be confused with periodicity. A periodic sequence repeats after a fixed number of steps, but it may take several different values. For example, $z_n=(-1)^n$ has period $2$, because $z_{n+2}=(-1)^{n+2}=(-1)^n=z_n$, yet $z_0=1$ and $z_1=-1$, so it is not constant. A constant sequence is periodic with every positive integer as a period, but a periodic sequence need not be constant.

It is also important to distinguish a constant sequence from an expression that only looks similar at a few indices. If two initial terms agree, that fact alone does not establish that all terms agree. A proof must use the entire defining formula or a rule that applies at every index, as in the induction argument above.

Check Your Understanding

Use the definition and the proved criteria to decide whether each sequence is constant, and identify its value when it is.

For $a_n=-3$ for every $n\in\mathbb{N}_0$, state the constant value and describe its range.
A sequence satisfies $b_{n+1}=b_n$ for every $n\in\mathbb{N}_0$, with $b_0=8$. Which theorem shows that the sequence is constant, and what is its value?
Is $c_n=\frac{5n+10}{n+2}$ constant for $n\in\mathbb{N}_0$? Simplify the formula and check that its denominator is nonzero.
If $a_n=4$ and $b_n=-3$ for every $n$, what are the constant values of $a_n+b_n$ and $a_nb_n$?
For $x_{n+1}=\frac{1}{2}x_n+5$, which equation must $x_0$ satisfy for the sequence to be constant? Does $x_0=10$ satisfy it?

What You'll Learn