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Monotone and Bounded Sequences · Tutorial 530 of 998

Proving Monotonicity With Ratios

Use the ratio of consecutive positive terms to prove monotonicity, including for sequences built from powers and factorials.

Intermediate 9 min read

What You'll Learn

  • State the ratio test for monotonicity of positive sequences.
  • Prove why comparing a consecutive-term ratio with one works.
  • Apply the test to factorial and power expressions.
  • Identify where a ratio changes from above one to below one.
  • Recognize why the positivity condition cannot be omitted.

Why Compare Consecutive Terms by a Ratio?

A sequence can be increasing or decreasing even when its terms involve powers, factorials, or both. In such cases, subtracting consecutive terms may produce an expression with many terms. If every term is positive, a ratio can give a cleaner comparison: it tells us how large the next term is relative to the current one.

For example, a ratio greater than one means that the next term is a factor larger than the current term. A ratio less than one means that the next term is a factor smaller. The method is exact, provided the sequence has positive terms.

Key idea: For a positive sequence, compare \(a_{n+1}/a_n\) with \(1\). A ratio above \(1\) indicates an increase from \(a_n\) to \(a_{n+1}\); a ratio below \(1\) indicates a decrease.

The Ratio Criterion

A sequence \((a_n)\) is nondecreasing if \(a_n\leq a_{n+1}\) for every \(n\in\mathbb{N}\), and it is nonincreasing if \(a_n\geq a_{n+1}\) for every \(n\in\mathbb{N}\). It is strictly increasing or strictly decreasing when the corresponding inequalities are strict for every \(n\).

Theorem: Suppose \(a_n>0\) for every \(n\in\mathbb{N}\). Then:
  • \((a_n)\) is nondecreasing if and only if \(a_{n+1}/a_n\geq 1\) for every \(n\).
  • \((a_n)\) is nonincreasing if and only if \(a_{n+1}/a_n\leq 1\) for every \(n\).
  • Strict inequalities in the ratio for every \(n\) give strict increase or strict decrease, respectively.

The criterion follows from the positivity of the denominator. Since \(a_n>0\), multiplying an inequality by \(a_n\) preserves its direction. Thus \(a_{n+1}/a_n\geq 1\) is equivalent to \(a_{n+1}\geq a_n\), and \(a_{n+1}/a_n\leq 1\) is equivalent to \(a_{n+1}\leq a_n\). The strict versions follow in the same way. Conversely, if the terms are ordered in either direction, division by the positive \(a_n\) preserves that order and gives the corresponding ratio inequality. This proves both directions of the criterion.

The ratio criterion therefore tests exactly the same consecutive-term comparison expressed by differences, but it can simplify expressions involving products, powers, and factorials. The method still requires checking the inequality for every index in the range being considered.

Worked Examples with Factorials and Powers

Worked Example: A Factorial Divided by a Power

Determine whether \(a_n=n!/2^n\) is monotone for \(n\in\mathbb{N}\). Here \(n!=1\cdot2\cdots n\), and \((n+1)!=(n+1)n!\). Every term is positive, so the ratio criterion applies.

Compute the ratio and simplify:

$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{2^{n+1}}\cdot\frac{2^n}{n!} = \frac{(n+1)n!}{2\cdot 2^n}\cdot\frac{2^n}{n!} = \frac{n+1}{2}. $$

For \(n=1\), this ratio is \(1\). For every \(n\geq2\), \(n+1\geq3\), so \((n+1)/2>1\). Thus the ratio is at least \(1\) for every \(n\), with equality only at the first step. The sequence is nondecreasing, and it is strictly increasing from \(a_2\) onward.

As a check, \(a_1=1/2\), \(a_2=2/4=1/2\), and \(a_3=6/8=3/4\). These agree with the ratio conclusions: the first two terms are equal, and the next term is larger.

Worked Example: A Factorial Divided by a Matching Power

Consider \(a_n=n!/n^n\) for \(n\in\mathbb{N}\). Again, every term is positive. Calculate the ratio using \((n+1)!=(n+1)n!\):

$$ \begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}\\ &= \frac{(n+1)n!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}\\ &= \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n. \end{aligned} $$

For every \(n\geq1\), \(0<n/(n+1)<1\). A positive integer power of a number strictly between zero and one is still strictly between zero and one. Therefore \(a_{n+1}/a_n<1\) for every \(n\), so the sequence is strictly decreasing.

Check the first steps: \(a_1=1\), \(a_2=2/4=1/2\), and \(a_3=6/27=2/9\). The first ratio is \(a_2/a_1=1/2\), matching \((1/2)^1\). The next ratio is \(a_3/a_2=(2/9)/(1/2)=4/9\), matching \((2/3)^2\). Both are below \(1\), as the general calculation requires.

When the Ratio Crosses One

A ratio need not stay on one side of \(1\). If it is above \(1\) for some indices and below \(1\) for others, the ratio criterion establishes the direction of change at each step. It may then show that the whole sequence is not monotone. The index range matters: a statement about all \(n\in\mathbb{N}\) differs from one about \(n\) beyond a specified threshold.

Worked Example: A Power Divided by a Factorial

Let \(a_n=3^n/n!\) for \(n\in\mathbb{N}\). All terms are positive. Simplifying the consecutive-term ratio gives

$$ \frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n} = \frac{3}{n+1}. $$

At \(n=1\), the ratio is \(3/2>1\), so the sequence increases from \(a_1\) to \(a_2\). At \(n=2\), the ratio is \(1\), so \(a_3=a_2\). For every \(n\geq3\), \(n+1\geq4\), giving \(3/(n+1)\leq3/4<1\); the sequence strictly decreases from then on.

This sequence is not monotone on all of \(\mathbb{N}\): its first step is an increase, while later steps are decreases. The terms confirm the calculations: \(a_1=3\), \(a_2=9/2\), \(a_3=27/6=9/2\), and \(a_4=81/24=27/8\). The ratios of these consecutive terms are \(3/2\), \(1\), and \(3/4\), respectively. They show an increase, then equality, then a decrease, exactly as predicted.

A Useful Pattern for Factorial Ratios

The preceding example illustrates a recurring simplification. For a fixed positive constant \(c\), consider \(a_n=c^n/n!\). The ratio is

$$ \frac{a_{n+1}}{a_n} = \frac{c^{n+1}}{(n+1)!}\cdot\frac{n!}{c^n} = \frac{c}{n+1}. $$

Thus the comparison with \(1\) reduces to comparing \(c\) with \(n+1\). When \(c/(n+1)>1\), the terms increase at that step; when it equals \(1\), the two terms are equal; and when it is less than \(1\), the terms decrease at that step. This calculation is a tool for a specified sequence, not a claim that every power-and-factorial sequence behaves the same way. The actual ratio must be computed for the sequence in question.

Common Pitfall: Forgetting Positivity

The ratio criterion depends on \(a_n>0\). If \(a_n=0\), the quotient \(a_{n+1}/a_n\) is undefined. If terms can be negative, dividing or multiplying an inequality by a term may reverse its direction. A ratio less than \(1\) is therefore not, by itself, evidence that an arbitrary sequence is decreasing.

For a concrete error, take \(a_n=(-1)^n\). Since \(a_{n+1}/a_n=-1<1\), one might incorrectly apply the positive-sequence rule and claim that the sequence is decreasing. But \(a_1=-1\) and \(a_2=1\), and \(-1<1\), so the first step is an increase. The terms continue to alternate, rather than decreasing. The correction is to check the hypothesis before using the rule: this sequence is not positive, so the ratio criterion stated here does not apply.

How to Use the Method Reliably

1
Check the terms.
Verify that \(a_n>0\) throughout the index range under consideration. This ensures the ratio is defined and comparisons preserve direction.
2
Form the next term.
Write \(a_{n+1}\) by replacing \(n\) with \(n+1\), taking care with exponents and factorials.
3
Simplify the ratio.
Compute \(a_{n+1}/a_n\) and cancel common factors before comparing the result with \(1\).
4
Compare over the full range.
Establish whether the ratio is above, equal to, or below \(1\) for every relevant index. If its behavior changes, state the index ranges and do not claim global monotonicity.

Careful bookkeeping is important. The ratio uses \(a_{n+1}\) in the numerator and \(a_n\) in the denominator. Reversing them reverses the comparison: \(a_n/a_{n+1}>1\), when defined and positive, corresponds to a decrease from \(a_n\) to \(a_{n+1}\), not an increase.

Why the Ratio Method Is Useful

Ratios often turn a complicated comparison into a simple expression. In factorial examples, the identity \((n+1)!=(n+1)n!\) creates a factor that cancels. For powers, \(c^{n+1}/c^n=c\). Together, these cancellations can leave a short expression whose comparison with \(1\) is clear. This is why the method is especially natural for sequences built from products, powers, and factorials.

The conclusion should match the evidence. A ratio at least \(1\) for every index proves nondecrease; a ratio at most \(1\) for every index proves nonincrease. If the inequalities are strict, the corresponding monotonicity is strict. If the ratio changes sides of \(1\), describe the behavior step by step and avoid assigning the sequence a monotonicity it does not have.

Check Your Understanding

For each question, use the ratio criterion only after checking its hypotheses.

  1. State the positivity condition needed to use \(a_{n+1}/a_n\) to prove monotonicity.
  2. If \(a_n>0\) and \(a_{n+1}/a_n=1\) for a particular index, what does that imply about the two consecutive terms?
  3. For \(a_n=4^n/n!\), compute \(a_{n+1}/a_n\) and determine the indices at which the sequence increases, stays equal, or decreases.
  4. Why does \(a_{n+1}/a_n<1\) not prove that a sequence with negative terms is decreasing?
  5. If a positive sequence has ratios above \(1\) for early indices and below \(1\) for later indices, what can you conclude about global monotonicity from those ratios?

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