Integration and series · Definition

Absolute convergence

real analysisseries

Before you start you need to know

It asks for more than convergence: the series must keep converging even when signs are ignored, and in return it behaves much more stably.

Definition

A series \(\sum a_n\) of real numbers converges absolutely when the series formed by the absolute values of its terms converges.

\[\sum_{n=1}^{\infty}|a_n|<\infty\]

The central fact is that absolute convergence implies convergence. If \(\sum|a_n|\) converges, its partial sums form a Cauchy sequence, and the triangle inequality \(|a_{m+1}+\dots+a_n|\le|a_{m+1}|+\dots+|a_n|\) makes the partial sums of \(\sum a_n\) Cauchy as well. By the completeness of \(\mathbb R\), the series converges, and moreover \(\left|\sum a_n\right|\le\sum|a_n|\).

The converse is false. The alternating harmonic series \(\sum(-1)^{n+1}/n\) converges (to \(\ln 2\)), but \(\sum 1/n\) diverges. A series that converges without converging absolutely is called conditionally convergent; its value depends on a delicate cancellation between positive and negative terms.

To decide absolute convergence, apply to \(\sum|a_n|\) any test for series of nonnegative terms, above all the comparison test: if \(|a_n|\le b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) converges absolutely. Thus \(\sum \sin(n)/n^2\) converges absolutely because \(|\sin n|/n^2\le1/n^2\).

The difference between the two kinds of convergence runs deep. In an absolutely convergent series the terms can be rearranged without changing the sum; in a conditionally convergent one, Riemann's theorem says a suitable rearrangement produces any prescribed sum, or even divergence. This is why absolute convergence is the natural notion when manipulating series.

Think about it

Give an example of a convergent series whose series of absolute values diverges.

The alternating harmonic series \(\sum(-1)^{n+1}/n\): it converges by Leibniz's test, but \(\sum 1/n\) diverges.

If \(\sum a_n\) converges absolutely, why is \(\left|\sum a_n\right|\le\sum|a_n|\)?

For each partial sum, \(\left|\sum_{k\le n}a_k\right|\le\sum_{k\le n}|a_k|\) by the triangle inequality; the inequality is preserved in the limit.

Exercises 10

References

  • Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §2.7, Teorema 2.7.6
  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 3, Teorema 3.45
  • Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §9.1, Definición 9.1.1