Sequences · Definition

Monotone sequence

real analysissequences

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.

\[a_n\le a_{n+1}\ \ \forall n\quad\text{or}\quad a_n\ge a_{n+1}\ \ \forall n\]

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

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