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The alternating series test with its error bound, and why a conditionally convergent series cannot be rearranged.
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 classify a series with varying signs as absolutely convergent, conditionally convergent or divergent, apply the alternating series test with its three hypotheses and read off its error bound, and say which manipulations of a series need absolute convergence. You will also be able to explain Riemann's rearrangement theorem, which is why the distinction is not a technicality.
The comparison, ratio and root tests, every one of which assumes non-negative terms or is really a test about sizes. This lesson is what happens when the signs are allowed to vary, and it is where a series stops behaving like a very long sum.
A series converges absolutely when the series of the sizes of its terms converges, and conditionally when it converges while that one does not. A rearrangement is the same terms written in a different order, which is a different series with, in general, a different sum.
$\sum a_n$ converges absolutely when $\sum |a_n|$ converges, and conditionally when it converges but $\sum |a_n|$ does not.
Absolute convergence implies convergence. By the Cauchy criterion for series, $\left|\sum_{n+1}^{m} a_k\right| \le \sum_{n+1}^{m} |a_k|$, and the right side is eventually below $\varepsilon$. So the tests of the last lesson, applied to the sizes, settle the original series too.
The alternating series test. If $b_n \ge 0$ decreases with $b_n \to 0$, then $\sum (-1)^{n+1} b_n$ converges, and the error after $n$ terms is at most $b_{n+1}$.
Why: the even partial sums increase, the odd ones decrease, every even one is below every odd one, and their gap is $b_{n+1} \to 0$. Two bounded monotone sequences closing on each other — monotone convergence, used twice.
What the distinction is for. An absolutely convergent series may be rearranged, split into two series and multiplied by another, all without further argument. A conditionally convergent one may not. Riemann's rearrangement theorem: the terms of a conditionally convergent series can be reordered to converge to any prescribed number, or to diverge.
So conditional convergence is not a minor blemish. It is the point at which a series stops being a sum in any ordinary sense, and the order of the terms becomes part of the answer.
Another way: picture
Plot the partial sums of an alternating series. They zigzag with shrinking swings, each one crossing the eventual sum, so the sum is always caught between the last two — and the width of that trap is the size of the next term, which is where the error bound comes from with no further work.
Another way: steps
To classify a series with varying signs:
Let $\sum a_n$ converge conditionally. Write $p_n$ for the positive terms and $q_n$ for the sizes of the negative ones.
If both $\sum p_n$ and $\sum q_n$ converged, the series of sizes would converge and the convergence would be absolute. If exactly one converged, the original series would diverge, since the other part is unbounded and unopposed. So both diverge.
That is the whole mechanism. Aim at any target $T$: take positive terms until the running total first exceeds $T$, then negative ones until it first falls below, and repeat. Each stage terminates because each part is unbounded, and the overshoot at each turn is at most one term, which tends to $0$. So the rearranged partial sums converge to $T$.
Nothing about the terms was used except that both halves diverge and the terms tend to zero, which is why the theorem is about every conditionally convergent series and not about a curious example.
| Operation | Needs | Fails without it |
|---|---|---|
| reordering the terms | absolute convergence | Riemann: any value reachable |
| splitting into two series | absolute convergence | both halves can diverge |
| multiplying two series | absolute convergence of one | the Cauchy product can diverge |
| grouping consecutive terms | convergence only | safe: this one is always allowed |
| dropping finitely many terms | nothing | safe: changes the sum, not the verdict |
The last two rows are worth noticing. Not every manipulation needs absolute convergence — inserting brackets around consecutive terms is always safe, because the bracketed partial sums are a subsequence of the original ones, and a subsequence of a convergent sequence converges to the same limit.
Applying the comparison test to terms that change sign. It assumes non-negative terms outright. Applied to a series with mixed signs it is not a weaker test; it is false.
Reading conditional convergence as almost-divergence. A conditionally convergent series has a sum, and its partial sums converge to it. What it lacks is the freedom to be rearranged.
Taking the error bound from the wrong term. The alternating bound is the size of the first term omitted, not the last one kept.
Using the alternating test without checking that the sizes decrease. Alternating signs and terms tending to zero are not enough; a series whose sizes decrease only on average can diverge, and the trapping argument is what fails.
For $\sum \dfrac{(-1)^n}{\sqrt{n}}$: the terms tend to $0$, so step 1 is passed.
The term test does not settle it.
The sizes are $1/\sqrt{n}$, a $p$-series with $p = 1/2 \le 1$, so $\sum |a_n|$ diverges: not absolute.
Step 2 rules out absolute convergence.
The signs alternate and $1/\sqrt{n}$ decreases to $0$, so the series converges conditionally, and stopping after $n$ terms has error at most $1/\sqrt{n+1}$.
Step 3 gives the verdict and the bound.
For $\sum \dfrac{\sin(n^2)}{n^2}$, the signs vary unpredictably and no alternating argument is available.
The terms have no usable pattern.
But $|a_n| \le 1/n^2$, and $\sum 1/n^2$ converges, so the series of sizes converges by comparison.
Bound the sizes and compare.
So the series converges absolutely, and may be rearranged at will. Nothing was ever established about where the signs fall.
Sizes settled it; the signs never mattered.
The terms tend to $0$, since $\ln n$ grows more slowly than $n$, so the term test does not settle it.
Step 1 of the recipe.
The sizes $\ln n / n$ exceed $1/n$ for $n \ge 3$, and $\sum 1/n$ diverges, so the series of sizes diverges: not absolute.
Comparison, in the right direction.
The sizes decrease from $n = 3$ onwards, since $\ln x / x$ decreases past $x = e$, so the alternating test applies from there and the series converges conditionally. Changing finitely many terms changes the sum and not the verdict, which is why starting the test late is legitimate.
Match each series to its verdict.
| converges absolutely | converges conditionally | diverges | |
|---|---|---|---|
| $\sum (-1)^n/n$ | |||
| $\sum (-1)^n/n^{4}$ | |||
| $\sum (-1)^n$ | |||
| $\sum (-1)^n n/(n+1)$ |
$S = \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k}$ is approximated by its partial sum $s_{9}$. What bound does the alternating series estimate give for $|S - s_{9}|$?
Answer:
Put the steps of the alternating series test, applied to $\sum \dfrac{(-1)^{n+1}}{n + 4}$, into order.
Number the steps in order (write the number in the box):
Here is an argument that the terms of $\sum \dfrac{(-1)^{n+1}}{n}$ may be reordered freely. Mark the one step that does not hold.
This task has no paper form; do it on a device.
$\sum (-1)^n/n^{5}$ converges absolutely and $\sum (-1)^n/n$ converges conditionally. What can be done with the first that cannot be done with the second?
For which real $x$ does $\displaystyle\sum_{n=0}^{\infty} (5x)^n$ converge?
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each series to its verdict.
| converges absolutely | converges conditionally | diverges | |
|---|---|---|---|
| $\sum (-1)^n/n$ | |||
| $\sum (-1)^n/n^{3}$ | |||
| $\sum (-1)^n$ | |||
| $\sum (-1)^n n/(n+1)$ |
You can classify a series with varying signs, use the alternating series test and its error bound, and say what absolute convergence licenses that conditional convergence does not. Say in your own words why commutativity of addition does not extend to an infinite series. Next: the same limit idea, now for a function of a real variable.
10. Your turn: classify $\sum \dfrac{(-1)^n \ln n}{n}$, step 3