Comparing Each Term with the Next
A sequence can be increasing or decreasing even when its terms are given by a formula that is not immediately easy to interpret. The central question is simple: for every index \(n\), how does \(a_{n+1}\) compare with \(a_n\)?
For an explicit sequence, a direct way to answer this is to calculate the difference \(a_{n+1}-a_n\). When every term is positive, a ratio can sometimes make the comparison quicker. The important task is to determine the sign of the comparison for every index in the sequence's domain, not just for a few sample terms.
The inequalities must hold throughout the specified index domain. A sequence that rises for a while and then falls is not monotone, even if most of its terms follow one direction. Strict monotonicity requires a strict inequality at every consecutive comparison; one equality is enough to rule out strict monotonicity.
The Consecutive-Difference Test
For each index \(n\), define the consecutive difference \(a_{n+1}-a_n\). Its sign tells you which term is larger:
- If \(a_{n+1}-a_n\geq 0\) for every \(n\), the sequence is increasing.
- If \(a_{n+1}-a_n>0\) for every \(n\), it is strictly increasing.
- If \(a_{n+1}-a_n\leq 0\) for every \(n\), the sequence is decreasing.
- If \(a_{n+1}-a_n<0\) for every \(n\), it is strictly decreasing.
If the difference has both positive and negative values as \(n\) varies, the sequence is neither increasing nor decreasing. Simplifying the difference algebraically is often more informative than calculating a long list of terms.
Worked Example: A Rational Sequence That Increases
Determine whether \(a_n=\frac{2n+1}{n+2}\), for \(n\in\mathbb{N}\), is monotone.
First substitute \(n+1\) for \(n\):
Now subtract the \(n\)th term. Since \(n\in\mathbb{N}\), both \(n+2\) and \(n+3\) are positive, so the denominator of the combined fraction is positive:
The numerator is positive and the denominator is positive for every \(n\in\mathbb{N}\), so \(a_{n+1}-a_n>0\) for every such \(n\). The sequence is strictly increasing.
As a check, \(a_1=\frac{3}{3}=1\) and \(a_2=\frac{5}{4}\), so \(a_2-a_1=\frac14\). The formula for the difference at \(n=1\) gives \(\frac{3}{4\cdot3}=\frac14\), in agreement.
Why Consecutive Comparisons Are Enough
Checking every consecutive pair determines comparisons between terms farther apart as well. The reason is that differences add: the intermediate terms cancel when consecutive differences are summed. This gives a complete justification for the test, not just a convenient procedure.
To see the cancellation directly, write out the sum:
Every intermediate term appears once with a positive sign and once with a negative sign. If each summand is nonnegative, their sum is nonnegative, so \(a_m-a_n\geq0\). If each summand is nonpositive, their sum is nonpositive, so \(a_m-a_n\leq0\). Thus the sign test for consecutive terms establishes the same order for any two terms in the sequence.
Using Ratios for Positive Sequences
When \(a_n>0\) for every \(n\), comparing \(a_{n+1}\) with \(a_n\) can also be done by comparing their ratio with \(1\). Dividing an inequality by a positive number preserves its direction. In particular, for positive \(a_n\), the conditions \(a_{n+1}>a_n\) and \(\frac{a_{n+1}}{a_n}>1\) are equivalent.
This is a comparison tool for sequences, not a general replacement for checking signs. If terms are zero, the ratio may be undefined. If terms are negative, dividing by them can reverse an inequality. In those cases, analyze \(a_{n+1}-a_n\) directly unless the signs and the direction of the comparison have been handled with care.
Worked Example: A Ratio Shows Nondecreasing Behavior
Determine whether \(a_n=\frac{2^n}{n}\), for \(n\in\mathbb{N}\), is increasing.
Every term is positive, so the ratio test applies. Compute:
Compare this ratio with \(1\):
For \(n\in\mathbb{N}\), the denominator is positive and \(n-1\geq0\). Therefore \(\frac{2n}{n+1}\geq1\) for every \(n\in\mathbb{N}\), with equality only when \(n=1\). The sequence is nondecreasing, and it is strictly increasing for \(n\geq2\). It is not strictly increasing over all of \(\mathbb{N}\), because \(a_1=a_2\).
Check the first terms: \(a_1=2\), \(a_2=2\), and \(a_3=\frac83\). The first comparison is equality, while the next is an increase, as the ratio analysis predicts.
When a Sequence Is Neither
To show that a sequence is not increasing, it is enough to find one index where \(a_{n+1}<a_n\). To show that it is not decreasing, it is enough to find one index where \(a_{n+1}>a_n\). If both types of comparison occur, the sequence is neither. A few terms can reveal such a failure, although a conclusion that a sequence is monotone requires proving the comparison for every index.
Worked Example: A Sequence That Is Neither Increasing nor Decreasing
Consider \(a_n=\frac{(-1)^n}{n}\), for \(n\in\mathbb{N}\). The first terms are:
The first difference is:
So the sequence is not decreasing. The next difference is:
So the sequence is not increasing. Since it has both an increase and a decrease between consecutive terms, it is neither increasing nor decreasing.
Check the arithmetic: \(-\frac13-\frac12=-\frac26-\frac36=-\frac56\), which is negative as required for the second comparison.
Common Pitfall: Using a Ratio Without Checking Signs
A ratio greater than \(1\) does not always mean that the next term is greater. The conclusion depends on the sign of the denominator term \(a_n\). For example, take \(a_n=-n\). The sequence begins \(-1,-2,-3,\ldots\), so it decreases. But its ratio is greater than \(1\):
The incorrect conclusion would be, “The ratio is greater than \(1\), so the sequence increases.” That reasoning silently assumes \(a_n>0\). Here \(a_n<0\), and multiplying the ratio inequality by \(a_n\) reverses the direction: \(a_{n+1}/a_n>1\) implies \(a_{n+1}<a_n\). Direct subtraction confirms the result:
Before using a ratio, check that every term in the denominator is positive. If that condition does not hold, the consecutive difference is usually the safer test.
What a Monotonicity Test Does—and Does Not—Establish
A successful sign test establishes the order of the terms. It does not, by itself, establish that the sequence has a finite limit. The Monotone Convergence Theorem, covered in the earlier material on that theorem, requires both monotonicity and boundedness for its convergence conclusion. Here the goal is to establish the monotonicity condition accurately, which is one part of that broader reasoning.
For an explicit sequence, a reliable approach is:
Replace \(n\) with \(n+1\) to obtain \(a_{n+1}\), keeping the sequence's index domain in view.
Use \(a_{n+1}-a_n\) in general. Use \(\frac{a_{n+1}}{a_n}\) only when all relevant terms are positive.
Simplify the comparison and check the signs of its numerator and denominator throughout the domain.
Distinguish increasing from strictly increasing, decreasing from strictly decreasing, and monotone from neither.
Testing a few terms is useful for checking arithmetic and spotting possible behavior, but it cannot replace the sign argument for all \(n\). The decisive evidence is a comparison formula whose sign has been established across the entire index domain.
Check Your Understanding
For each question, justify your conclusion using consecutive differences or, when applicable, ratios.
- For \(a_n=3n+2\), is the sequence increasing, strictly increasing, decreasing, or neither?
- For \(b_n=\frac{1}{n+1}\), determine whether the sequence is strictly monotone. Show the sign of \(b_{n+1}-b_n\).
- For \(c_n=\frac{3^n}{n}\), compute \(\frac{c_{n+1}}{c_n}\) and determine its relation to \(1\) for \(n\in\mathbb{N}\).
- Why is a ratio-based conclusion about monotonicity unsafe when the terms may be negative?
- What two different signs among consecutive differences would show that a sequence is neither increasing nor decreasing?
