This week is devoted to understanding the convergence of series. Among convergent series, there is a distincting between two kinds of convergences: a stronger and a weaker version. This section explores that distinction via the important example of the alternating harmonic series. Let me start with a definition.
Definition 10.2.1.
Let \(\{a_n\}\) be a sequence such that \(a_n>0 \ \forall n \in \NN\text{.}\) the following series is called an alternating series.
\begin{equation*}
\sum_{n=1}^\infty (-1)^{n+1} a_n
\end{equation*}
Since the \(a_n\) are positive, the \((-1)^n\) ensure that each consecutive term in the series changes sign, from positive to negative back to positive. This is the ‘alternating’ that gives the series its name.
Now recall the test for divergence in
Proposition 9.2.11. From that test, I observe that for a series to converge, it is
necessary but not
sufficient for the terms to tend to zero. Intuitively, I would be much more convenient if this condition was sufficient, but the harmonic series was the counter example. For alternating series, it turns it is sufficient.
Proposition 10.2.2.
(The Alternating Series Test) An alternating series converges if and only if the limit of the terms is zero.
I won’t give a formal proof, the I’ll describe the informal idea. In an alternating series, in each step of the sum I alternatively add and subtract term. The sum is always increasing when I add and then decreasing when I subtract. It’s always jumping forward and then jumping back down.
In that jumping forward and jumping back, if the terms are getting smaller, then the jumps are getting smaller as well. Therefore, if the terms go to zero, that’s enough to ensure that this foward and backward pattern will eventually convergence on some value.
Subsection 10.2.1 The Alternating Harmonic Series
To further understand the convergence of alternative series, I’ll work with a specific and important example.
Definition 10.2.3.
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 of the series is \(\ln 2\text{.}\) However, very strange things happen when I try to re-arrange the terms of the harmonic series. Consider the following series.
\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*}
I’m going to do some strange arithmetic here. I can express this series as the following sum. I think about adding the two series term-wise, adding each term in the second row to the term above it. Hopefully, you can see that doing so will cancel half the terms and add up to exactly the terms of the previous series in the othter terms.
\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 \hspace{0.2cm} + 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 I factor \(\frac{1}{2}\) out of the second series, it is also the harmonic series. I 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? Let me simply list the positive and negative term in order. 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 I 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.
Proposition 10.2.4.
For any real number \(\alpha\text{,}\) there is a re-arrangement of the alternating harmonic series that sums to \(\alpha\text{.}\)
Proof.
This is a very strange result, but the proof has a remarkably simple argument. First, groups the terms as positive and negative, as I did before. Each set of terms is asymptotically similar to the (non-alternating) harmonic series, so each set sums to \(\pm \infty\) (or \(-\infty\) for the negative terms).
Now let \(\alpha\) be any real number. I will start by adding positive terms until I get past \(\alpha\text{.}\) (This can always be done, since the positive terms by themslves sum to \(\infty\)). Then, when the sum so far is larger than \(\alpha\text{,}\) I start adding negative terms until I get back below \(\alpha\) again. (Again, this can always be done, since the negative terms sum to \(-\infty\)). Then I simply repeat this process, adding positives until I get above \(\alpha\) and negatives until I get back below \(\alpha\text{.}\) This process can be continued indefinitely. Since both the positive and negative terms get arbitrarily small, this process will invariably converge on \(\alpha\) in the limit.
Example 10.2.5.
There are some regular arrangements of the alternating harmonic which have specific values. Let \(A(m,n)\) be the sum formed by adding \(m\) positive terms followed by \(n\) negative terms and repeating this pattern. (As before, the positive and negative terms are taken in decreasing order.) It can be proved that this converges to certain values of the logarithm.
\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 \hspace{0.5cm} + \frac{1}{3} - \frac{1}{10} -
\frac{1}{12} - \frac{1}{14} - \frac{1}{16}\\
\amp \hspace{0.5cm} + \frac{1}{5} - \frac{1}{18} -
\frac{1}{20} - \frac{1}{22} - \frac{1}{24}\\
\amp \hspace{0.5cm} + \frac{1}{7} - \frac{1}{26} -
\frac{1}{28} - \frac{1}{30} - \frac{1}{32}\\
\amp \hspace{0.5cm} + \frac{1}{9} - \frac{1}{34} -
\frac{1}{36} - \frac{1}{38} - \frac{1}{40} \ldots
\end{align*}
Subsection 10.2.2 Conditional Convergence
This situation for the alternating harmonic series is not unique.
Definition 10.2.6.
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 I demonstrated for the alternating harmonic series is the same for any conditionally convergent series.
Proposition 10.2.7.
An absolutely convergent series converges to the same value regardless of re-ordering, but a conditionally convergent series can be rearranged to converge to any real number.
