Absolute and conditional convergence

4 Absolute and conditional convergence

The ratio test applies to series of positive terms. Indeed this is true of many related tests for convergence. However, as we have seen, not all series are series of positive terms. To apply the ratio test such series must first be converted into series of positive terms. This is easily done. Consider two series $\sum_{p = 1}^{\infty} a_{p}$ and $\sum_{p = 1}^{\infty} |a_{p}|$ . The latter series, obviously directly related to the first, is a series of positive terms.

Using imprecise language, it is harder for the second series to converge than it is for the first, since, in the first, some of the terms may be negative and cancel out part of the contribution from the positive terms. No such cancellations can take place in the second series since they are all positive terms. Thus it is plausible that if $\sum_{p = 1}^{\infty} |a_{p}|$ converges so does $\sum_{p = 1}^{\infty} a_{p}$ . This leads to the following definitions.

Key Point 8

Conditional Convergence and Absolute Convergence

A convergent series $\sum_{p = 1}^{\infty} a_{p}$ is said to be conditionally convergent if $\sum_{p = 1}^{\infty} |a_{p}|$ is divergent.

A convergent series $\sum_{p = 1}^{\infty} a_{p}$ is said to be absolutely convergent if $\sum_{p = 1}^{\infty} |a_{p}|$ is convergent.

For example, the alternating harmonic series:

$\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p + 1}}{p} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$

is conditionally convergent since the series of positive terms (the harmonic series):

$\sum_{p = 1}^{\infty} |\frac{{(- 1)}^{p + 1}}{p}| \equiv \sum_{p = 1}^{\infty} \frac{1}{p} = 1 + \frac{1}{2} + \frac{1}{3} + \dots$

is divergent.

Task!

Show that the series $- \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \dots$ is absolutely convergent.

First, find the general term of the series:

$- \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \dots = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p}}{(2 p)!} ∴ a_{p} \equiv \frac{{(- 1)}^{p}}{(2 p)!}$

Write down an expression for the related series of positive terms:

$\sum_{p = 1}^{\infty} \frac{1}{(2 p)!} so a_{p} = \frac{1}{(2 p)!}$

Now use the ratio test to examine the convergence of this series:

$p^{t h}$ term $= \frac{1}{(2 p)!}$ ${(p + 1)}^{t h}$ term $= \frac{1}{(2 (p + 1))!}$

Find $lim_{p \to \infty} [\frac{{(p + 1)}^{t h} term}{p^{t h} term}]$ :

$\frac{(2 p)!}{(2 (p + 1))!} = \frac{2 p (2 p - 1) \dots}{(2 p + 2) (2 p + 1) 2 p (2 p - 1) \dots} = \frac{1}{(2 p + 2) (2 p + 1)} \to 0$ as $p$ increases.

So the series of positive terms is convergent by the ratio test. Hence $\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p}}{(2 p)!}$ is absolutely convergent.

Exercises

Which of the following alternating series are convergent?
1. $\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p} ln (3)}{p}$
2. $\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p + 1}}{p^{2} + 1}$
3. $\sum_{p = 1}^{\infty} \frac{p sin (2 p + 1) \frac{π}{2}}{(p + 100)}$
Use the ratio test to examine the convergence of the series:
1. $\sum_{p = 1}^{\infty} \frac{e^{4}}{{(2 p + 1)}^{p + 1}}$
2. $\sum_{p = 1}^{\infty} \frac{p^{3}}{p!}$
3. $\sum_{p = 1}^{\infty} \frac{1}{\sqrt{p}}$
4. $\sum_{p = 1}^{\infty} \frac{1}{{(0.3)}^{p}}$
5. $\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p + 1}}{3^{p}}$
For what values of [maths rendering] are the following series absolutely convergent?
1. $\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p} x^{p}}{p}$
2. $\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p} x^{p}}{p!}$

1. convergent,
2. convergent,
3. divergent
1. $λ = 0$ so convergent
2. $λ = 0$ so convergent
3. $λ = 1$ so test is inconclusive. However, since $\frac{1}{p^{1 ∕ 2}} > \frac{1}{p}$ then the given series is divergent by comparison with the harmonic series.
4. $λ = 10 ∕ 3$ so divergent, (e) Not a series of positive terms so the ratio test cannot be applied.
1. The related series of positive terms is $\sum_{p = 1}^{\infty} \frac{{|x|}^{p}}{p}$ . For this series, using the ratio test we find $λ = |x|$ so the original series is absolutely convergent if $|x| < 1$ .
2. The related series of positive terms is $\sum_{p = 1}^{\infty} \frac{{|x|}^{p}}{p!}$ . For this series, using the ratio test we find $λ = 0$ (irrespective of the value of $x$ ) so the original series is absolutely convergent for all values of $x$ .

4 Absolute and conditional convergence

Key Point 8

Task!

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Exercises

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