Why Compare Consecutive Terms?
A recursive rule defines terms using earlier terms. To determine whether such a sequence is increasing or decreasing, the most direct question is: how does the next term compare with the current one? Subtracting the current term from the next term often turns the recurrence into an expression whose sign can be analyzed.
This comparison is useful because it works with the recurrence as written. The difference may simplify, factor, or reveal a condition on earlier terms. But the sign must be justified: a formula for the difference does not by itself establish that the sequence is monotone for every index.
The Basic Comparison
Suppose a sequence is defined by a one-step recurrence \(x_{n+1}=F(x_n)\). The difference to examine is obtained by subtracting \(x_n\) from both sides:
If this expression is nonnegative for every term \(x_n\) that occurs, the sequence is nondecreasing. If it is nonpositive for every term that occurs, the sequence is nonincreasing. The phrase “that occurs” matters: a sign condition on \(F(t)-t\) over some interval is useful only if the sequence’s terms are known to lie in that interval.
For a recurrence that uses two previous terms, such as \(x_{n+1}=G(x_n,x_{n-1})\), the same subtraction gives
The resulting expression may depend on both earlier terms. The task is then to establish its sign using conditions that are actually known about those terms.
Why the Difference Determines Monotonicity
The comparison criterion follows directly from the order properties of the real numbers. For any two real numbers \(a\) and \(b\), \(b-a\ge 0\) is equivalent to \(b\ge a\). Applying this with \(a=x_n\) and \(b=x_{n+1}\) gives the exact link between the difference and the term-by-term order.
Express \(x_{n+1}\) in terms of the earlier terms specified by the definition.
Form \(x_{n+1}-x_n\), taking care to subtract the entire expression for \(x_n\).
Factor, combine fractions, or use known conditions on the earlier terms.
If the sign holds for every relevant index, conclude monotonicity; otherwise, state only the comparison that has been justified.
This method does not require the recurrence function \(F\) to be increasing. It compares \(F(x_n)\) directly with \(x_n\). The sign of \(F(t)-t\), rather than the way \(F\) compares two different inputs, is the information being used here.
Worked Examples
Worked Example: A Difference That Is Always Positive
Let \(x_1=1\) and define \(x_{n+1}=x_n^2+1\). Determine whether the sequence is increasing.
Subtract \(x_n\) from the recurrence:
Complete the square to determine the sign for every real value of \(x_n\):
A square is nonnegative, and \(3/4\) is positive, so the difference is strictly positive at every index. Thus the sequence is strictly increasing.
Check the first terms: \(x_1=1\), \(x_2=1^2+1=2\), \(x_3=2^2+1=5\), and \(x_4=5^2+1=26\). The consecutive differences are \(2-1=1\), \(5-2=3\), and \(26-5=21\). The formula gives \(1^2-1+1=1\), \(2^2-2+1=3\), and \(5^2-5+1=21\), confirming the calculation.
Worked Example: A Two-Step Recurrence
Let \(x_1=0\), \(x_2=1\), and define \(x_{n+1}=x_n+x_{n-1}^2+1\) for \(n\ge 2\). Compare consecutive terms.
Subtract \(x_n\) from the recurrence:
For every real \(x_{n-1}\), its square is nonnegative, so \(x_{n-1}^2+1>0\). Therefore \(x_{n+1}>x_n\) for every \(n\ge 2\). Along with \(x_2=1>x_1=0\), this proves that the sequence is strictly increasing.
Check the initial terms and differences: \(x_3=1+0^2+1=2\); \(x_4=2+1^2+1=4\); and \(x_5=4+2^2+1=9\). The differences from \(x_1\) onward are \(1-0=1\), \(2-1=1\), \(4-2=2\), and \(9-4=5\). The difference formula gives \(x_1^2+1=1\), \(x_2^2+1=2\), and \(x_3^2+1=5\) for the comparisons starting at \(n=2\), in agreement with the computed terms.
Worked Example: A Negative Difference with a Positive Recurrence
Let \(x_1=1\) and define \(x_{n+1}=\dfrac{x_n}{1+x_n^2}\). Determine whether the sequence is decreasing.
Subtract \(x_n\) and combine the terms over a common denominator:
The initial term is positive. If \(x_n>0\), then \(1+x_n^2>0\) and the recurrence gives \(x_{n+1}>0\). Thus every term is positive, so the numerator \(-x_n^3\) is negative and the denominator is positive. Consequently \(x_{n+1}-x_n<0\) at every index, and the sequence is strictly decreasing.
For a numerical check, \(x_2=1/(1+1)=1/2\), \(x_3=(1/2)/(1+1/4)=2/5\), and \(x_4=(2/5)/(1+4/25)=10/29\). The differences are \(1/2-1=-1/2\), \(2/5-1/2=-1/10\), and \(10/29-2/5=-8/145\). The difference formula at \(n=3\) gives \(- (2/5)^3/(1+(2/5)^2)=-(8/125)/(29/25)=-8/145\), as required.
Sign Conditions and Common Pitfalls
Sometimes the difference does not have one sign for all possible real inputs. In that case, identify the condition that makes the sign definite, and check whether that condition is known for the terms of the sequence. For example, a factor \(x_n(x_n-1)\) is nonnegative when \(x_n\le 0\) or \(x_n\ge 1\), and negative when \(0<x_n<1\). Without knowing which range contains \(x_n\), this factor alone does not establish whether the sequence is increasing or decreasing.
A frequent error is to confuse a positive next term with a positive difference. A term can be positive while being smaller than the term before it. The comparison must use \(x_{n+1}-x_n\), not just the sign of \(x_{n+1}\).
The correct difference is
The terms begin \(x_1=0\), \(x_2=2-0=2\), \(x_3=2-2=0\), and \(x_4=2-0=2\). The differences are \(2\), \(-2\), and \(2\), so the sequence is neither nondecreasing nor nonincreasing. In particular, the sign of the next term is not the sign test for monotonicity.
Another error is to establish a sign only for a few calculated terms and extend it to all indices. Computations can check an algebraic formula or illustrate a pattern, but a monotonicity conclusion requires a sign argument that applies at every relevant index.
When This Method Is Useful
Consecutive-term comparison is especially effective when the recurrence produces a simple difference after subtraction. A square plus a positive constant, a product of factors with known signs, or a fraction with a denominator known to be positive can make the direction of change clear. With a recurrence involving more than one earlier term, the same idea applies, but the sign may depend on those terms and their allowed values.
The essential discipline is to separate the algebra from the sign reasoning. First derive the exact expression for \(x_{n+1}-x_n\). Then state why each factor or denominator has the sign claimed, and specify what is known about the sequence terms. This makes clear both what the recurrence establishes and what still needs justification.
Check Your Understanding
For each question, focus on the difference between consecutive terms and the conditions needed to determine its sign.
- If \(x_{n+1}=x_n^2+4\), what expression equals \(x_{n+1}-x_n\), and is its sign fixed for every real \(x_n\)?
- For \(x_{n+1}=x_n+x_{n-1}^2\), write the consecutive difference and state when it is strictly positive.
- Why does knowing \(x_{n+1}>0\) not, by itself, imply that \(x_{n+1}>x_n\)?
- For a recurrence \(x_{n+1}=F(x_n)\), which expression must be analyzed to compare consecutive terms?
- If a difference formula has a denominator, what must be checked before using the numerator’s sign to determine the fraction’s sign?