Sequences · Definition
Monotone sequence
Before you start you need to know
Monotonicity is the simplest order condition on a sequence and, together with boundedness, it is enough to guarantee convergence.
Definition
A real sequence \((a_n)\) is monotone when, as the index \(n\) grows, its terms either never decrease or never increase.
A sequence \((a_n)\) is increasing if \(a_n\le a_{n+1}\) for all \(n\), and decreasing if \(a_n\ge a_{n+1}\) for all \(n\). When the inequalities are strict we speak of strictly increasing or strictly decreasing sequences. A constant sequence is both increasing and decreasing in the weak sense.
There are three usual ways to check monotonicity of \(a_n\): study the sign of \(a_{n+1}-a_n\), compare the quotient \(\frac{a_{n+1}}{a_n}\) with \(1\) when the terms are positive, or differentiate a function \(f\) with \(a_n=f(n)\). For instance, \(a_n=\frac{n}{n+1}\) satisfies \(a_{n+1}-a_n=\frac1{(n+1)(n+2)}>0\), so it is strictly increasing.
Two mistakes are common. The first is to believe that «monotone» requires every term to change: the sequence \(1,1,2,2,3,3,\dots\) is increasing. The second is to think the first term decides the direction: if \(a_1=-4\) and \(a_{n+1}=\frac{a_n}2\), the terms are negative and the sequence grows toward \(0\). Nor should monotone be confused with bounded: \(a_n=n\) is monotone and unbounded, and \((-1)^n\) is bounded and not monotone.
The sum of increasing sequences is increasing, but the product, difference and quotient need not be. A non-monotone sequence may still converge (like \(\frac{(-1)^n}{n}\)), yet monotonicity gives a very strong convergence criterion in the next concept: the monotone convergence theorem.
Think about it
Is the sequence \(a_n=\frac{(-1)^n}{n}\) monotone?
No: its terms \(-1,\ \tfrac12,\ -\tfrac13,\ \tfrac14,\dots\) alternate in sign, so they go up and down in turn.
If \(a_n\) is increasing and \(b_n=-a_n\), what can be said about \(b_n\)?
It is decreasing, because multiplying by \(-1\) reverses inequalities: \(a_n\le a_{n+1}\) implies \(-a_n\ge-a_{n+1}\).
Exercises 10
- The sequence \(a_n=2n+5\) is:
- The sequence \(a_n=\frac1n\) is:
- The sequence \(a_n=(-1)^n\) is:
- The sequence \(a_n=\frac{n}{n+1}\) is:
- Let \(a_n=n^2-6n\). Which statement is correct?
- Let \(a_1=-4\) and \(a_{n+1}=\frac{a_n}{2}\). The sequence is:
- If \((a_n)\) is increasing and \(a_1=3\), what always holds?
- The sequence \(a_n=\frac{2n}{n^2+1}\) is:
- If \((a_n)\) and \((b_n)\) are increasing, which sequence is always increasing?
- Let \(a_n=\frac{n}{2^n}\). Which statement is correct?
What comes next
References
- Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §2.4
- Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 3, 3.13
- Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §3.3