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Alternating signs let $\sum \tfrac{(-1)^n}{n}$ converge where $\sum \tfrac1n$ cannot; three conditions are needed and the decreasing one is the one that gets skipped. Then a second question — what $\sum|a_n|$ does — splits convergence into absolute and conditional.
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 can apply the alternating series test with all three of its conditions stated, and classify a convergent series as absolutely or conditionally convergent by testing $\sum|a_n|$ separately. You can say why a divergent absolute series proves nothing on its own, and what a conditionally convergent sum loses: the right to be added up in any order.
Every test so far has needed positive terms: comparison, the integral test and the $p$-rule all break down the moment signs are allowed to change. This lesson is about the commonest way signs do change — strict alternation — and about the new question that opens up once they do.
Alternating series: one whose terms change sign every time, usually written with a factor $(-1)^n$.
The alternating series test: if the signs alternate, the sizes $|a_n|$ decrease, and $|a_n| \to 0$, the series converges.
Absolutely convergent: $\sum |a_n|$ converges.
Conditionally convergent: $\sum a_n$ converges but $\sum |a_n|$ does not.
The test. If the terms alternate in sign, their sizes $|a_n|$ decrease, and $|a_n| \to 0$, then the series converges.
Three conditions, and the middle one is the one that gets skipped. The reason they work together is worth seeing once: each partial sum overshoots the answer and the next undershoots it, by less each time, so the partial sums close in on a single number like a shrinking pair of jaws.
Alternation is powerful. $\sum \tfrac1n$ diverges and $\sum \tfrac{(-1)^n}{n}$ converges, on identical sizes of terms. The signs are doing all the work: the cancellation between consecutive terms is what keeps the partial sums from running away.
Which raises a second question. A series can converge because of cancellation, or it can converge because its terms are genuinely small. Those are different situations and they are told apart by looking at $\sum |a_n|$:
Absolute convergence implies convergence, so testing $\sum |a_n|$ first is often the fastest route — and when it converges you are finished in one step. When it diverges you have learned nothing about the original and must still apply the alternating test.
Why the distinction is drawn. An absolutely convergent series can be added up in any order and gives the same total. A conditionally convergent one can be rearranged to give any total you choose, or none at all — so its sum belongs to the order as much as to the numbers.
Another way: picture
Picture stepping forward one metre, back half a metre, forward a third, back a quarter. You overshoot the destination, then undershoot it, then overshoot by less; the target is always trapped between where you are and where you will be next. Now picture the same step lengths all forward: you walk past every landmark there is. The sizes are identical and only the turning round makes the journey finite.
Another way: steps
| $\sum a_n$ | $\sum \lvert a_n\rvert$ | Verdict | The sum is |
|---|---|---|---|
| converges | converges | absolutely convergent | independent of the order |
| converges | diverges | conditionally convergent | a property of the order too |
| diverges | diverges | divergent | not a number |
There is no fourth row: a series whose absolute version converges cannot itself diverge. So a convergent absolute series settles everything at once, and a divergent one settles nothing on its own.
Skipping the decreasing condition. The test needs the sizes to decrease, not merely to tend to zero. Write the check down; on a marked paper it is worth a mark of its own.
Concluding divergence from a divergent absolute series. $\sum \left|\tfrac{(-1)^n}{n}\right|$ diverges and $\sum \tfrac{(-1)^n}{n}$ converges. The absolute series can prove convergence and can never prove divergence.
Treating 'converges' as the whole answer. When a question says determine whether the series converges absolutely, converges conditionally, or diverges, it is asking for one of three named outcomes, and the classification is part of what is being marked.
$\displaystyle\sum \frac{(-1)^{n}}{\sqrt{n}}$: the sizes $\tfrac{1}{\sqrt n}$ decrease and tend to zero, and the signs alternate.
All three conditions hold.
So it converges. Now the absolute series: $\sum \tfrac{1}{\sqrt n}$ is a $p$-series with $p = \tfrac12$, which diverges.
A separate question.
So the convergence is conditional — it depends on the cancellation, not on the terms being small.
$\displaystyle\sum \frac{(-1)^{n}}{n^{2}}$: try the absolute series first, $\sum \tfrac{1}{n^2}$.
The fast route.
It is a $p$-series with $p = 2 > 1$, so it converges.
One line.
Absolute convergence implies convergence, so the original converges absolutely and the alternating test was never needed.
Look at the sizes first: $\dfrac{n}{2n+1} \to \tfrac12$, not to zero.
Always the first check.
So the third condition of the alternating test fails, and the $n$th-term test applies instead.
Alternation cannot rescue this.
The series diverges. No amount of sign-changing helps a series whose terms do not shrink to nothing.
Match each alternating series to its classification and the reason for it.
| Conditionally: without the signs it is the harmonic series | Absolutely: without the signs it is a $p$-series with $p = 2$ | Diverges: the sizes of the terms do not reach zero | Absolutely: without the signs it is geometric | |
|---|---|---|---|---|
| $\sum \dfrac{(-1)^n}{n}$ | ||||
| $\sum \dfrac{(-1)^n}{n^2}$ | ||||
| $\sum \dfrac{(-1)^n n}{n+1}$ | ||||
| $\sum \dfrac{(-1)^n}{2^n}$ |
For each series, say what the series of absolute values does and then classify the original.
| Without the signs it | So the series converges | |
|---|---|---|
| $\sum \dfrac{(-1)^n}{n}$ | ||
| $\sum \dfrac{(-1)^n}{n^{5}}$ | ||
| $\sum \dfrac{(-1)^n n}{n + 1}$ |
Classify $\sum \dfrac{(-1)^n}{n}$: does it converge absolutely, converge conditionally, or diverge?
Put the steps of classifying $\displaystyle\sum \dfrac{(-1)^{n}}{n^{5}}$ into the order you do them.
Number the steps in order (write the number in the box):
Select every condition the alternating series test actually requires.
This task has no paper form; do it on a device.
The alternating harmonic series converges conditionally. Its terms are now written down in a different order, and the new series is added up. What can the total be?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Classify $\sum \dfrac{(-1)^n}{n}$: does it converge absolutely, converge conditionally, or diverge?
You can classify an alternating series three ways. Without looking: what are the three conditions of the alternating series test, and what does a divergent series of absolute values establish on its own?
9. Your turn: $\displaystyle\sum \frac{(-1)^{n}\,n}{2n + 1}$, step 3