Convergência condicional

We've learned that a series $\sum a_{n}$ absolutely converges if the series $\sum |a_{n}|$ converges (and hence $\sum a_{n}$ converges). The other possibility for a convergent series also has a name:

Definition. A series $\sum a_{n}$ which converges, but which $\sum |a_{n}|$ diverges, is called a conditionally convergent series.

Let's look at a few examples.

Conditionally convergent series have a very strange property: remember when there were vague warnings earlier on about how it can sometimes be `illegal' to rearrange the terms of a series when determining its convergence or divergence, or its value when convergent?

It turns out that a conditionally convergent series can always be arranged in such a way so that the rearranged series is either divergent (to $\infty$ or $-\infty$, if we choose), or convergent to any limit we want. What the?!

As weird as this sounds, it is actually quite easy to (intuitively) explain how this is possible. We'll demonstrate how this is done with the alternating harmonic series, and the technique easily generalizes to any conditionally convergent series.

First, we'll separate the positive and negative terms of the alternating harmonic series. The positive terms are given by

$$1, \frac{1}{3}, \frac{1}{5}, \ldots.$$

Let's call this sequence $a_{n}$. The even terms are given by

$$-\frac{1}{2}, -\frac{1}{4}, -\frac{1}{6}, \ldots;$$

let's call this sequence $b_{n}$.

Notice that the series formed by just the positive terms diverges to $\infty$, whereas the series formed by just the even terms diverge to $-\infty$. One consequence is that given any starting point in $a_{n}$, and some arbitrary (positive) number $L$, we can sum consecutive terms from that starting point and eventually exceed $L$. For example, if $L = 1$ and we start at $1/1001$, we can find some $n$ such that

$$\frac{1}{1001} + \frac{1}{1003} + \ldots + \frac{1}{2n+1} > 1.$$

(A calculator / computer program shows that the smallest such $2n+1$ is $7389$.)

Suppose we now want to rearrange the terms of the alternating harmonic series to make this rearranged series diverge to $\infty$. We will construct this rearrangement in the following iterative way.

At step $k$, we will sum together whatever positive terms we have left until the sum we are constructing exceeds $k$. Then we will subtract the $k$th negative term. We continue this process indefinitely; it is obvious that this is a rearrangement of the alternating harmonic series. In addition, by the way we are making this construction, the partial sums are tending towards $\infty$.

Let's make this more concrete with a few examples. At step 1, we want to sum enough positive terms to exceed 1, so we'll take $1 + 1/3$. We then subtract the first negative term, $1/2$. At the end of step 1, the partial sum we have constructed is

$$1 + \frac{1}{3} - \frac{1}{2}.$$

At step 2, we sum enough of the remaining positive terms to get this sum to exceed 2. This is possible because of the basic fact we mentioned earlier. In this case, the smallest number of positive terms we can add is

$$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} + \ldots + \frac{1}{41} \approx 2.004.$$

We then subtract $1/4$ from this. We can keep continuing in this fashion; it is clear the partial sums are (slowly) tending towards $\infty$, hence the rearranged series tends towards $\infty$.

It isn't hard to generalize this method to obtain any positive limit $L$ we want. At step $k$ of this iterative procedure, we now sum enough positive terms to just exceed $L$, and then we sum enough negative terms to just go under $L$.

Again, let's look at a specific example with the alternating harmonic series and $L = 7$. (We could pick any $L$ and what we're about to do would still work! That's what makes this so amazing!) We start by summing enough positive terms to just exceed 7; a computer program shows this is achieved by

$$1 + \frac{1}{3} + \ldots + \frac{1}{337607} \approx 7.00000179178.$$

We subtract $1/2$ from this to obtain

$$1 + \ldots + \frac{1}{337607} - \frac{1}{2} \approx 6.50000179178.$$

We then add enough remaining positive terms to just exceed 7; this is achieved by

$$1 + \ldots + \frac{1}{337607} - \frac{1}{2} + \frac{1}{337609} + \ldots + \frac{1}{917711} \approx 7.000000870176198.$$

So forth and so on; notice that the partial sums are (very slowly) converging to 7, because the negative term we subtract at step $k$ is equal to $-1/2k$, which tends towards 0.

If we want to get negative limits instead of positive limits, we just swap the role of positive and negative terms in the above iterative procedure. Or if we want, we could rearrange a conditionally convergent series to diverge, but not to $\pm \infty$. (For example, we could get the partial sums to continually oscillate between 0 and 1.)

Proving that the above technique works for a general conditionally convergent series is slightly harder, but the principle is identical. In particular, the two key properties we use are the facts that the series consisting of the positive terms (1) diverges to $\infty$, and (2) the $n$th term converges to 0, and similarly for the series consisting of negative terms, with $\infty$ replaced by $-\infty$.

(TODO: fill in a rigorous proof of the above.)