Why Compare Consecutive Terms?
A list of initial terms can suggest that a sequence increases or decreases, but a few calculations do not establish what happens for every index. To prove monotonicity, we need a comparison that works for all relevant \(n\). For an explicit sequence, a direct method is to subtract consecutive terms and determine the sign of the result.
This method turns a question about the order of infinitely many terms into an algebra problem: simplify \(a_{n+1}-a_n\), then determine whether it is always nonnegative, always nonpositive, or changes sign.
For a sequence indexed by \(\mathbb{N}=\{1,2,3,\dots\}\), the consecutive difference is defined by \(\Delta a_n=a_{n+1}-a_n\), for \(n\in\mathbb{N}\). The sign of this difference records the order of the two terms: a positive difference means the next term is larger, a negative difference means it is smaller, and a zero difference means the terms are equal.
Why the Difference Test Works
The result follows directly from subtracting the same quantity from both sides of an inequality. This is not a shortcut that merely predicts the order: it is an equivalent way to write the definition of monotonicity.
For nondecreasing behavior, we need \(a_{n+1}\ge a_n\) for every index \(n\).
Subtracting \(a_n\) from both sides preserves the inequality, giving \(a_{n+1}-a_n\ge 0\).
If \(a_{n+1}-a_n\ge 0\), adding \(a_n\) to both sides gives \(a_{n+1}\ge a_n\). The same reversible argument applies to nonincreasing sequences and to strict inequalities.
Thus, to prove a sequence is nondecreasing, it is enough to calculate \(\Delta a_n\) and establish that it is nonnegative for every index in the domain. For a nonincreasing sequence, establish that it is nonpositive for every index. The words “for every index” matter: checking a few values can be useful for spotting a pattern, but it does not prove the sign condition holds throughout the domain.
Worked Example: A Polynomial Sequence That Changes Direction
Worked Example: A Polynomial Sequence That Changes Direction
Determine whether \(a_n=n^2-4n\), for \(n\in\mathbb{N}\), is monotone.
First substitute \(n+1\) for \(n\):
Now subtract \(a_n\):
At \(n=1\), the difference is \(2(1)-3=-1\), which is negative. For every \(n\ge 2\), \(2n-3\ge 2(2)-3=1>0\). Therefore, the sequence decreases from its first term to its second, and then increases at every subsequent step.
In particular, it is not nondecreasing, because \(a_2-a_1=-1<0\); it is not nonincreasing, because \(a_3-a_2=1>0\). As a check, \(a_1=1-4=-3\), \(a_2=4-8=-4\), and \(a_3=9-12=-3\). These values agree with the differences \(-1\) and \(1\). The sign analysis proves the behavior for all indices, not just these terms.
Rational Sequences: Track the Denominator
For rational expressions, combining terms usually produces a fraction for \(\Delta a_n\). Its sign depends on both numerator and denominator. A useful approach is to establish the denominator’s sign on the full domain, then analyze the numerator. If the denominator is positive, the fraction has the same sign as its numerator. If its sign is not known, it must be considered before drawing a conclusion.
Worked Example: A Rational Sequence That Is Strictly Increasing
Show that \(a_n=\frac{n}{n+1}\), for \(n\in\mathbb{N}\), is strictly increasing.
Calculate the next term and subtract:
For every \(n\in\mathbb{N}\), both \(n+1\) and \(n+2\) are positive. The denominator is therefore positive, so \(a_{n+1}-a_n>0\) for every \(n\). By the difference criterion, the sequence is strictly increasing.
As a check at the first index, \(a_1=\frac12\) and \(a_2=\frac23\), so \(a_2-a_1=\frac23-\frac12=\frac16\). The general formula gives \(\frac{1}{(3)(2)}=\frac16\), as required.
Sometimes the numerator of a rational difference is not always positive or always negative. In that case, its sign may vary with \(n\). One zero or sign change can matter: a sequence is nondecreasing only if every consecutive difference is nonnegative, and nonincreasing only if every one is nonpositive.
Worked Example: A Rational Sequence That Is Not Monotone
Determine whether \(a_n=\frac{n}{n^2+4}\), for \(n\in\mathbb{N}\), is monotone.
Both denominators in the consecutive terms are positive, since \(n^2+4>0\) and \((n+1)^2+4>0\). Compute the difference using a common denominator:
The denominator is positive for every \(n\in\mathbb{N}\), so the sign is determined by \(4-n^2-n\). At \(n=1\), this numerator is \(4-1-1=2>0\). For every \(n\ge2\), \(n^2+n\ge 2^2+2=6\), so \(4-n^2-n\le -2<0\). The sequence increases from \(a_1\) to \(a_2\), then decreases at every later step. It is therefore neither nondecreasing nor nonincreasing.
Check the first two differences directly: \(a_1=\frac15\), \(a_2=\frac{2}{8}=\frac14\), and \(a_3=\frac{3}{13}\). Thus \(a_2-a_1=\frac14-\frac15=\frac1{20}>0\), while \(a_3-a_2=\frac3{13}-\frac14=\frac{12-13}{52}=-\frac1{52}<0\). These agree with the general sign analysis.
A Common Algebraic Pitfall
The most important practical danger is an error in substituting \(n+1\) or distributing a minus sign. Because a sign error can reverse the conclusion, keep the parentheses until each operation has been applied carefully.
A second common error with rational expressions is to inspect the numerator and ignore the denominator. Before deciding the sign of a fraction, state why its denominator is positive or negative on the entire index domain. If it can be zero, the sequence itself may not be defined at that index, so the domain must be addressed first.
A Reliable Procedure
For an explicit sequence, use the following sequence of steps. This keeps the algebra and the logical claim aligned.
Identify the indices for which both \(a_n\) and \(a_{n+1}\) are defined.
Replace every occurrence of \(n\) in the formula by \(n+1\), preserving parentheses.
Calculate \(\Delta a_n=a_{n+1}-a_n\), distributing the subtraction across the entire expression for \(a_n\).
For fractions, check the denominator as well as the numerator. Identify any indices where the difference is zero or changes sign.
Say whether the sequence is strictly or weakly increasing or decreasing, or explain why it is not monotone.
The method matters because monotonicity is a statement about every pair of consecutive terms in the domain. An algebraic sign argument can establish that universal comparison directly. Numerical terms can check a calculation or suggest what to investigate, but they cannot replace the sign analysis for all indices.
Check Your Understanding
For each question, use the consecutive difference and justify its sign over the stated domain.
- For \(a_n=3n^2+n\), \(n\in\mathbb{N}\), calculate \(a_{n+1}-a_n\) and determine whether the sequence is strictly increasing.
- For \(b_n=\frac{1}{n+1}\), \(n\in\mathbb{N}\), calculate the consecutive difference and classify the sequence.
- Why must the denominator sign be checked when analyzing a rational consecutive difference?
- If \(a_{n+1}-a_n\) is negative at one index and positive at another, what can you conclude about whether \((a_n)\) is monotone?
- For \(c_n=n^2-3n\), \(n\in\mathbb{N}\), find the indices where the consecutive difference is negative, zero, or positive, and describe the resulting behavior.