Back to the on-screen lesson ·
Two questions with three possible answers, and a rearrangement theorem that makes the difference between them matter.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to test the series of absolute values first and say why that order saves work, prove that absolute convergence implies convergence and see why the converse fails, classify any series as absolutely convergent, conditionally convergent or divergent, explain why the fourth combination cannot occur, describe how a conditionally convergent series is rearranged to reach any target, and say which manipulations each kind of convergence permits.
The ratio and root tests concluded that $\sum|a_n|$ converges — absolute convergence. Leibniz's test concluded only that $\sum a_n$ converges, and said nothing about the sizes.
That difference was quietly recorded and not explained. It matters, because the two conclusions are genuinely different: $\sum \frac{(-1)^{n+1}}{n}$ converges, to $\ln 2$, while $\sum \frac1n$ diverges. This lesson is about what separates them, and about what the weaker kind of convergence fails to guarantee.
$\sum a_n$ converges absolutely if $\sum |a_n|$ converges. It converges conditionally if it converges but $\sum|a_n|$ does not.
Those are the only two kinds of convergence, and with divergence they make three classifications. The fourth combination — $\sum|a_n|$ convergent and $\sum a_n$ not — cannot happen, and the theorem that rules it out is the main one here.
A rearrangement of a series is the same terms in a different order, each used exactly once. Rearranging a finite sum never changes it; rearranging an infinite one can, and whether it does is exactly the absolute/conditional distinction.
Theorem. If $\sum|a_n|$ converges then $\sum a_n$ converges.
Proof. $0 \le a_n + |a_n| \le 2|a_n|$, so $\sum(a_n + |a_n|)$ converges by comparison — a series of non-negative terms, where all the earlier machinery applies. Then $\sum a_n = \sum(a_n + |a_n|) - \sum|a_n|$ is a difference of convergent series.
That is what makes the ratio and root tests usable on series with mixed signs: they bound the sizes, and this theorem converts that into convergence.
The converse fails. $\sum\frac{(-1)^{n+1}}{n}$ converges; $\sum\frac1n$ does not. So there are three classifications and not two.
The procedure. Given any series:
Why the distinction matters: rearrangement. If $\sum a_n$ converges absolutely, every rearrangement converges to the same sum. If it converges conditionally, Riemann's rearrangement theorem says its terms can be reordered to converge to any real number you choose, or to diverge.
The mechanism is that in a conditionally convergent series the positive terms alone sum to $+\infty$ and the negatives alone to $-\infty$. Each half is an unlimited supply, so a greedy construction can chase any target. In an absolutely convergent series both halves are finite and no such freedom exists.
Another way: picture
Two buckets, one collecting the positive terms and one the negative. For an absolutely convergent series both buckets fill to finite levels and the sum is their difference — a fixed number, whatever order you pour in. For a conditionally convergent one both buckets are bottomless, and the sum you see is an artefact of how you alternated between them.
Another way: steps
Riemann's theorem is usually stated and not shown, which makes it sound like a pathology. The construction is short and worth following once.
Take $\sum\frac{(-1)^{n+1}}{n}$ and a target, say $2$. The positive terms $1, \tfrac13, \tfrac15, \ldots$ sum to $+\infty$, and the negatives $-\tfrac12, -\tfrac14, \ldots$ to $-\infty$.
Add positive terms, in order, until the running total first exceeds $2$. This must happen, since the positives are unlimited. Then add negative terms until the total first drops below $2$. Then positives again until it exceeds $2$, and so on forever.
Every term is used exactly once, so this is a rearrangement. And at each turning point the overshoot is at most the size of the last term used — which tends to zero. So the running totals converge to $2$.
Nothing about $2$ was special, and the same construction reaches any target, or runs off to $+\infty$ if you never turn back. What made it possible was that both halves diverge. In an absolutely convergent series the positives sum to some finite $P$ and the negatives to $-N$, every rearrangement gives $P - N$, and there is no freedom to exploit.
Reporting "converges" when "converges absolutely" is true. Not wrong, but weaker than what you proved, and the stronger statement is what permits rearranging and multiplying series.
Testing the original series first. If the absolute series converges, the original does too, and one test has settled both questions. Going the other way round can mean doing two.
Assuming an alternating series converges conditionally. It may converge absolutely — $\sum\frac{(-1)^n}{n^2}$ does — and it may diverge.
Applying Leibniz to decide absolute convergence. Leibniz says nothing about $\sum|a_n|$, which is a series of positive terms and needs a positive-term test.
Rearranging or regrouping a conditionally convergent series. Not permitted, and the error is invisible: you get a number, and it is the wrong one.
Thinking conditional convergence is rare or artificial. It is the normal situation for alternating series built from $p$-series with $p \le 1$, and those are common.
Addition is commutative. Every learner has known that since primary school, and it is true of every finite sum. The natural extension — that an infinite sum is determined by which numbers are being added — is false, and it is false in the strongest possible way: for a conditionally convergent series the answer can be made anything at all.
What has gone wrong is that the sum was never an addition. It is the limit of the sequence of partial sums, and reordering the terms produces a different sequence of partial sums, which may have a different limit. The terms are a set; the partial sums are a sequence; and the limit belongs to the sequence.
Absolute convergence is exactly the condition under which that distinction stops mattering. When both the positive part and the negative part are finite, every ordering gives the same two totals and hence the same difference. When both are infinite, the visible sum is a record of how you alternated between them.
So the classification is not bookkeeping. It tells you which manipulations are permitted: rearranging, regrouping, multiplying two series together, integrating term by term. Absolute convergence permits them; conditional convergence does not. Reporting which kind you have is part of the answer, and the reason it is part of the answer is that somebody later will want to do one of those things.
$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}n}{2^{n}}$: the signs alternate, but take absolute values first.
The absolute series is the one to test.
$\sum \dfrac{n}{2^{n}}$ has ratio $\dfrac{n+1}{2n} \to \dfrac12 < 1$, so it converges.
A positive-term test on the positive series.
So the original converges absolutely — and one test did both questions. Leibniz would also have given convergence, but only the weaker kind, and would have taken as long.
$\displaystyle\sum_{n=2}^{\infty}\frac{(-1)^{n}}{\sqrt{n}\,}$: absolute values give $\sum n^{-1/2}$, a $p$-series with $p = \tfrac12 \le 1$, which diverges.
The first question, answered no.
So test the original separately. Signs alternate; sizes $n^{-1/2}$ decrease and tend to zero; Leibniz applies.
The second question, answered yes.
Conditionally convergent. So its sum may be changed by reordering — and any manipulation that reorders or regroups the terms is off limits until that is noticed.
Absolute values: $\sum \dfrac{\ln n}{n}$. Its terms exceed $\dfrac1n$ for $n \ge 3$, and $\sum \frac1n$ diverges, so it diverges.
Above a divergent series is the useful direction.
So test the original. Sizes $\dfrac{\ln n}{n}$ tend to $0$; do they decrease? Differentiating $\dfrac{\ln x}{x}$ gives $\dfrac{1 - \ln x}{x^{2}}$, negative for $x > e$, so yes from $n = 3$ onwards.
The condition that has to be proved rather than assumed.
Leibniz applies from $n = 3$, and the first two terms are a finite sum. So the series converges conditionally.
Classify $\sum \dfrac{(-1)^n}{2^{n}}$.
Match each series to its classification.
| Converges absolutely | Converges conditionally | Diverges | |
|---|---|---|---|
| $\sum \dfrac{(-1)^{n}}{n^{2}}$ | |||
| $\sum \dfrac{(-1)^{n}}{n}$ | |||
| $\sum \dfrac{(-1)^{n}\,n}{n + 6}$ | |||
| $\sum \dfrac{(-1)^{n}}{7^{\,n}}$ |
For each series, answer both questions.
| Converges? | Converges absolutely? | |
|---|---|---|
| $\sum \dfrac{(-1)^{n}}{n^{4}}$ | ||
| $\sum \dfrac{(-1)^{n}}{n}$ | ||
| $\sum (-1)^{n}$ |
Put the steps of classifying $\displaystyle\sum\frac{(-1)^{n}}{n + 6}$ into the order you do them.
Number the steps in order (write the number in the box):
In $\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}$ the positive terms are $1, \tfrac13, \tfrac15, \ldots$. How many of the first $16$ terms of the series are positive?
Answer:
$\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \ln 2$. Can its terms be reordered so that the sum is $6$ instead?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Classify $\sum \dfrac{(-1)^n}{2^{n}}$.
You can classify a series of mixed signs into one of three kinds and say what each kind permits. Say in your own words why the sum of a conditionally convergent series depends on the order of its terms. Next: series with a variable in them, where convergence becomes a question about which values of $x$.
10. Your turn: classify $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}\ln n}{n}$, step 3