Skip to main content

Section 1.2 Absolute and Conditional Convergence

Among convergent series, there is a distincting between a stronger and a weaker kind of convergence. This section explores that distinction via the important example of the alternating harmonic series.

Subsection 1.2.1 Alternating Series Test

Definition 1.2.1.

If we have a sequences of terms \(\{a_n\}\) such that \(a_n>0 \ \forall n \in \NN\text{,}\) then the following expression is called an alternating series.

\begin{equation*} \sum_{n=1}^\infty (-1)^{n+1} a_n \end{equation*}

In an alternating series, each term has a different sign from the previous term. Recall the test for divergence: for convergence, is it necessary but not sufficient for the terms to tend to zero. Intuitively, we would like to have sufficiency as well, but the harmonic series was the counter example. For alternating series, we get our wish.

Subsection 1.2.2 The Alternating Harmonic Series

Definition 1.2.3.

The alternating harmonic series is the harmonic series with \((-1)^n\) in the numerator.

\begin{equation*} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \end{equation*}

This series converges by the alternating series test. It is difficult to prove, but the value is \(\ln 2\text{.}\)

We must be very careful here. Consider the following series, which is a pattern of two positive terms with odd denominators followed by one negative term with an even denominator.

\begin{align*} \amp 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + \ldots\\ \end{align*}

We write this series in a strange new way: as the difference of the alternating harmonic series and another series with only even denominators. You can see, if you add the two series, how some of the terms cancel and other add to the correct terms in the original.

\begin{align*} \amp = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \ldots\\ \amp + 0 + \frac{1}{2} + 0 - \frac{1}{4} + 0 + \frac{1}{6} + 0 - \frac{1}{8} + \ldots\\ \end{align*}

The first series is the harmonic series. If we factor \(\frac{1}{2}\) out of the second series, it is also the harmonic series. We use the known value of \(\ln 2\) for the harmonic series to calculate the value of this series.

\begin{align*} \amp = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} + \frac{1}{2} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \ln 2 + \frac{1}{2} \ln 2 = \frac{3}{2} \ln 2 \end{align*}

This looks reasonable as well, but what are the terms of this series? If we group them by sign, the positive terms are \(\{ 1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots \}\) and the negative terms are \(\{ \frac{-1}{2}, \frac{-1}{4}, \frac{-1}{6}, \ldots \}\text{.}\) These are exactly the same terms at the alternating harmonic series, just in a different order. However, the alternating harmonic series summed to \(\ln 2\text{,}\) not \(\frac{3}{2} \ln 2\text{.}\)

It seems we can re-arrange the alternating harmonic series to sum to a different number. This is exceedingly odd: for finite sums, any re-arrangement was irrelevant to the value of the sum. It seems, for infinite sums, re-arrangement can actually change the value. There is an important result which is even stranger.

This is a very strange result, but the proof has a remarkably simple argument.First, groups the terms as positive and negative. Each set of terms is asymptotically similar to the (non-alternating) harmonic series, so each set sums to \(\pm \infty\text{.}\)

Then choose a real number \(\alpha\text{.}\) Start adding positive terms until we get past \(\alpha\text{.}\) (This can always be done, since the possitive terms by themslves sum to \(\infty\)). Then, when we are past \(\alpha\text{,}\) start adding negative terms until we're below \(\alpha\) again. (Again, this can always be done, since the negative terms sum to \(-\infty\)). Then simply repeat this process, adding positives until we get above \(\alpha\) and negatives until we get back below \(\alpha\text{.}\) This process can be continued indefinitely, and since the terms get arbitrarily small, we will approach \(\alpha\) in the limit.

There are some regular arrangements of the alternating harmonic which have specific values. Let \(A(m,n)\) be the sum where we take \(m\) positive terns, then \(n\) negative, then back to \(m\) positive and so on. It can be proved that this converges to

\begin{equation*} A(m,n) = \ln 2 + \frac{1}{2} \ln \left( \frac{m}{n} \right)\text{.} \end{equation*}

In particular, the combination of one positive and four negative terms sums to zero.

\begin{align*} A(1,4) \amp = \ln 2 + \frac{1}{2} \ln \frac{1}{4} = \ln 2 + \ln \left( \frac{1}{4} \right)^{\frac{1}{2}} = \ln 2 + \ln \frac{1}{2} = \ln 2 - \ln 2 = 0\\ 0 \amp = 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \frac{1}{8}\\ \amp + \frac{1}{3} - \frac{1}{10} - \frac{1}{12} - \frac{1}{14} - \frac{1}{16}\\ \amp + \frac{1}{5} - \frac{1}{18} - \frac{1}{20} - \frac{1}{22} - \frac{1}{24}\\ \amp + \frac{1}{7} - \frac{1}{26} - \frac{1}{28} - \frac{1}{30} - \frac{1}{32}\\ \amp + \frac{1}{9} - \frac{1}{34} - \frac{1}{36} - \frac{1}{38} - \frac{1}{40} \ldots \end{align*}

Subsection 1.2.3 Conditional Convergence

This situation for the alternating harmonic series is not unique.

Definition 1.2.6.

A convergent series \(\sum a_n\) is called absolutely convergent if

\begin{equation*} \sum_{n=1}^\infty |a_n| \lt \infty\text{.} \end{equation*}

Otherwise, if a series is convergent but not absolutely convergent, it is called conditionally convergent.

The alternating harmonic series was a conditionally convergent series, since the (non-alternating) harmonic series diverges. The behaviour that we saw for the alternating harmonic series is the same for any conditionally convergent series.