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Leibniz's two conditions on the sizes make the odd and even partial sums close in from both sides, and the error is the first term left out.
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 state Leibniz's three conditions and check each of them separately, prove that the sizes decrease by algebra or by differentiating the matching function rather than assuming it, follow the nested-interval proof and say which hypothesis each step uses, quote the error bound as the size of the first omitted term and solve it for the number of terms a given accuracy needs, recognise that a failed test is not a verdict of divergence, and say why the conclusion is not absolute convergence.
The integral test, the comparison tests and the argument behind the ratio and root tests all rest on the same fact: for non-negative terms the partial sums increase, so convergence is boundedness.
With mixed signs that collapses. The partial sums can go up and down, and boundedness no longer implies convergence — $1 - 1 + 1 - 1 + \cdots$ has partial sums bounded between $0$ and $1$ and no limit. A different mechanism is needed, and for the special case where the signs alternate strictly, Leibniz found one: the partial sums close in on the answer from both sides at once.
A series is alternating if consecutive terms have opposite signs, every time. It is usually written $\sum(-1)^{n}b_n$ or $\sum(-1)^{n+1}b_n$ with $b_n \ge 0$.
The sizes are the $b_n$: the terms with the signs stripped off. Everything the test asks about is a statement about the sizes.
Leibniz's conditions are that the sizes decrease and that they tend to zero. Both are needed.
The remainder $R_N = S - s_N$ is the error made by stopping after $N$ terms. The remarkable conclusion of the test is that $|R_N| \le b_{N+1}$ — the size of the very next term.
Alternating series test (Leibniz). If $b_n \ge 0$, $b_{n+1} \le b_n$ for all $n$, and $b_n \to 0$, then $$\sum_{n=1}^{\infty}(-1)^{n+1}b_n$$ converges.
The error bound comes with it. $$|S - s_N| \le b_{N+1},$$ and moreover $S$ lies strictly between $s_N$ and $s_{N+1}$. This is the only test in the course whose conclusion hands you an error estimate for free.
Why it is true. The even partial sums increase, the odd ones decrease, and every even one is below every odd one — so the two sequences are monotone and bounded and each converges. Their limits differ by $\lim b_n = 0$, so they agree, and the partial sums converge to the common value. The picture is two sequences of nested intervals closing on a point.
Both conditions are needed. Without decreasing, the nesting fails and the test is false — there are alternating series with sizes tending to zero non-monotonically that diverge. Without tending to zero, the divergence test already ends it.
Decreasing eventually is enough. Convergence depends on the tail, so the sizes may misbehave for the first few terms; apply the test from where the decrease starts.
This test proves convergence only. If it fails, the series may still converge for another reason, or diverge. And the convergence it proves may be conditional — the subject of the next lesson.
Another way: picture
A number line with the partial sums marked in order: $s_1$ far right, $s_2$ to the left of it, $s_3$ between them, $s_4$ between $s_2$ and $s_3$, and so on. Each new term jumps a shorter distance than the last and in the opposite direction, so the marks spiral inwards on a point. The sum is that point, and it is always between the last two marks — which is the error bound, read off the picture.
Another way: steps
For a series of positive terms, the error after $N$ terms is the whole tail, and bounding it takes work — an integral, or a geometric estimate. For an alternating series satisfying Leibniz's conditions it is at most one term.
The reason is cancellation. The tail is $b_{N+1} - b_{N+2} + b_{N+3} - \cdots$, and because the sizes decrease, grouping it as $(b_{N+1} - b_{N+2}) + (b_{N+3} - b_{N+4}) + \cdots$ shows it is non-negative, while grouping it as $b_{N+1} - (b_{N+2} - b_{N+3}) - \cdots$ shows it is at most $b_{N+1}$. Trapped between $0$ and $b_{N+1}$, and with the sign of the first omitted term.
The practical difference is large. For $\sum \frac{(-1)^{n+1}}{n^2}$, ten terms give an error below $\frac{1}{121}$. For $\sum \frac{1}{n^2}$, ten terms leave a tail of about $\frac{1}{10}$ — twelve times worse, from the same sizes. The alternation is doing that.
It also explains why the bound requires the sizes to be decreasing, not merely small. The grouping argument uses $b_{k} \ge b_{k+1}$ at every step; without it the tail can be larger than its first term, and the bound is simply false.
Taking alternation as the hypothesis. It is one of three conditions and the least consequential. $\sum(-1)^n \frac{n}{n+1}$ alternates and diverges.
Not checking that the sizes decrease. This is the condition that is usually assumed and sometimes false. $b_n = \frac{2 + (-1)^n}{n}$ tends to zero and does not decrease monotonically.
Concluding divergence when the test fails. The test proves convergence and nothing else. If it does not apply, nothing has been established.
Bounding the error by the tail rather than by the first omitted term. The point of the test is that the sharper bound is available.
Reading the conclusion as absolute convergence. It is not. $\sum \frac{(-1)^{n+1}}{n}$ converges and $\sum\frac1n$ does not, which is the subject of the next lesson.
Applying it to a series that is not strictly alternating. Terms whose signs follow some other pattern — two positive, one negative — are not covered, whatever their sizes do.
The test is named for the alternation, so the alternation gets the attention, and a great many answers consist of "the signs alternate, so it converges". That sentence is wrong about every series whose terms do not shrink, and there are plenty.
Of the three conditions, alternation is the cheapest to check and the least restrictive. Decreasing is the one that makes the nesting work, and it is the one that is often assumed without proof — sometimes wrongly, since "tends to zero" does not imply "decreases". Tending to zero is required by the divergence test anyway, so a series failing it was never a candidate.
What alternation buys is cancellation, and the value of cancellation is visible in the error bound. A positive series' tail is the sum of everything left; an alternating one's tail is smaller than its first term, because each term is partly undone by the next. That is the whole mechanism, and it is why the conditions are what they are: the terms must be shrinking for each to be undone by less than itself.
So when writing up a Leibniz argument, give the three conditions equal billing and spend the effort on the second. "The signs alternate, the sizes decrease (because $f'(x) \le 0$ for $x \ge 2$), and the sizes tend to zero" is a complete argument. Anything shorter is usually missing the part that could fail.
$\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n^{2}+1}$: the signs alternate and the sizes tend to zero. Do they decrease?
The third condition is the one to check.
It is not obvious from the formula, so differentiate the matching function: $f(x) = \dfrac{x}{x^2+1}$ has $f'(x) = \dfrac{1 - x^2}{(x^2+1)^2}$, which is $\le 0$ for $x \ge 1$.
Calculus, used on the sizes.
So the sizes decrease from $n = 1$ onwards and the test applies: the series converges. The absolute series behaves like $\sum \frac1n$ and diverges, so the convergence here is conditional.
$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{3n + 1}{2n + 5}$: the signs alternate.
The easy condition.
The sizes tend to $\dfrac32$, not to zero, so Leibniz does not apply.
The test declines.
And the divergence test settles it: terms whose sizes approach $\frac32$ do not tend to zero, so the series diverges. Alternation never rescues terms that fail to shrink — the partial sums oscillate by about $\frac32$ forever.
Signs alternate; sizes are $\dfrac{1}{\ln n}$, which decrease because $\ln$ increases, and tend to $0$ because $\ln n \to \infty$.
All three conditions, checked.
So the series converges by Leibniz.
Convergence, but not absolute: $\sum \frac{1}{\ln n}$ has terms larger than $\frac1n$ and diverges.
The error after the terms up to $n = 11$ is at most $\dfrac{1}{\ln 12} \approx 0.40$ — a poor bound, because $1/\ln n$ shrinks extremely slowly. Convergent and useless for computation is a perfectly ordinary combination.
Does the alternating series test settle $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n + 3}$?
$\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}$ satisfies Leibniz's conditions. Bound the error made by stopping after $7$ terms.
Answer:
Put the steps of applying the alternating series test to $\displaystyle\sum(-1)^{n}\frac{1}{n + 8}$ into the order you do them.
Number the steps in order (write the number in the box):
For $\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}$, fill in the first four partial sums.
| Value | |
|---|---|
| $s_1$ | 1 |
| $s_2$ | |
| $s_3$ | |
| $s_4$ |
Build the proof that if $b_n$ decreases to $0$ then $\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}b_n$ converges.
This task has no paper form; do it on a device.
How many terms of $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}$ are needed to be sure of an error below $\dfrac{1}{16}$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Does the alternating series test settle $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n + 3}$?
You can settle an alternating series and attach an error bound to a truncation of it. Say in your own words why the error is one term rather than a whole tail. Next: the difference between a series that converges and one that converges absolutely.
10. Your turn: does $\displaystyle\sum_{n=2}^{\infty}\frac{(-1)^{n}}{\ln n}$ converge, and what is the error after $10$ terms?, step 3