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If $\lim a_n \ne 0$ the series diverges; if $\lim a_n = 0$ the test says nothing at all. The cheapest check in the unit, the one most often quoted backwards, and the harmonic series as the standing reason it cannot run the other way.
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 $n$th-term test: find the limit of a general term, compare it with zero, and conclude divergence when it misses. You can also say precisely what the test concludes when the limit is zero — nothing — and why it cannot be strengthened, using the harmonic series and $\sum n^{-2}$ as the pair of series that forces the silence.
You can take the limit of a sequence, and you know that a series converges when its partial sums settle. This lesson connects the two in the only way they can be connected cheaply — and then spends most of its effort on how little that connection gives you.
The $n$th-term test, also called the divergence test: if $\lim a_n \ne 0$, then $\sum a_n$ diverges.
Inconclusive: the test returns no information, so the series is exactly as undecided as before.
Necessary condition: something every convergent series must have. Terms tending to zero is one. Sufficient condition: something that would be enough on its own. Terms tending to zero is not one.
If a series converges, its terms must tend to zero. The reason is short: $a_n = s_n - s_{n-1}$, and if the partial sums approach a limit $L$ then both $s_n$ and $s_{n-1}$ approach $L$, so their difference approaches $L - L = 0$.
Turn that around and you have a test:
$$\text{if } \lim_{n\to\infty} a_n \ne 0 \text{ (or does not exist), then }\sum a_n \text{ diverges.}$$
That is the whole of it, and it only runs that way. The converse — terms to zero, therefore convergent — is false, and the harmonic series $\sum \tfrac1n$ is the standing counter-example: its terms go to zero and its partial sums grow without limit.
Terms going to zero is necessary and nowhere near sufficient. The harmonic series has terms going to zero and its sum is infinite; that single example is why the $n$th-term test is stated in one direction only and can never certify convergence.
Why bother with a test that usually says nothing? Because it is free, and because when it does speak it ends the problem in one line. Checking it first costs a few seconds and occasionally saves a page. On an examination, a series whose terms visibly tend to something other than zero is a mark you should not spend two minutes on.
Another way: picture
Picture a bank account you keep paying into. If the deposits do not shrink to nothing, the balance cannot settle — that much is obvious, and it is the whole content of the test. But deposits shrinking to nothing does not mean the balance settles: deposit a penny less each day for ever and you still pay in an unbounded amount. Small is not the same as small enough.
Another way: steps
| The limit of the terms | What the test concludes |
|---|---|
| Not zero | the series diverges |
| Does not exist | the series diverges |
| Zero | nothing at all |
There is no row of this table whose conclusion is converges, and there never can be: the harmonic series and $\sum n^{-2}$ both have terms tending to zero, and one diverges while the other converges. A test that spoke in the third row would have to give the same verdict to two series with different verdicts.
Reading the test backwards. The terms go to zero, so it converges is the single most common false step in this unit. The test has no such branch.
Writing 'converges' where 'inconclusive' belongs. On a marked paper these are not the same answer even when the series happens to converge, because the reasoning given does not support the conclusion drawn.
Skipping the test because it usually says nothing. It is the cheapest check available and it is the only one that can finish a problem before it starts. Run it first, every time, and move on quickly when it is silent.
$\displaystyle\sum \frac{3n + 2}{4n - 1}$: divide top and bottom by $n$ to get $\dfrac{3 + 2/n}{4 - 1/n}$.
The standard move.
That tends to $\tfrac34$, which is not zero.
The hypothesis holds.
So the series diverges, and no other test is needed. The terms are eventually all close to $\tfrac34$, so the partial sums climb by about that much for ever.
$\displaystyle\sum \frac{1}{\sqrt{n}}$: the terms tend to zero.
Slowly, but they get there.
So the $n$th-term test is inconclusive. It has no opinion on this series.
Not a verdict of convergence.
The series in fact diverges, and the next two lessons give the tools that show it. Silence from this test is not evidence either way.
The term is $\cos(1/n)$, and $\tfrac1n \to 0$, so the argument of the cosine goes to zero.
Take the limit of the term, not of the series.
$\cos 0 = 1$, so $a_n \to 1$, which is not zero.
The inner limit is zero; the term's limit is not.
So the series diverges by the $n$th-term test. The trap here is reading the $\tfrac1n$ inside and concluding the term goes to zero.
Match each term to the limit it tends to as $n$ grows.
| $1$ | $\tfrac23$ | $3$ | $0$ | |
|---|---|---|---|---|
| $\dfrac{n}{n + 1}$ | ||||
| $\dfrac{2n + 1}{3n - 1}$ | ||||
| $\dfrac{3n}{n + 2}$ | ||||
| $\dfrac{1}{n}$ |
A series has terms $a_n = \dfrac{1}{n}$. What does the $n$th-term test conclude?
For each term, give its limit and say what the $n$th-term test concludes.
| Limit | The test concludes | |
|---|---|---|
| $\dfrac{n}{n + 1}$ | ||
| $\dfrac{1}{n^2}$ | ||
| $\dfrac{4n + 1}{n + 4}$ |
Find $\displaystyle\lim_{n\to\infty} \dfrac{4n + 8}{3n + 9}$, which is the first thing the $n$th-term test needs.
Answer:
Put the steps of applying the $n$th-term test to $\displaystyle\sum \dfrac{5n}{n + 3}$ into the order you do them.
Number the steps in order (write the number in the box):
Select every series below that the $n$th-term test settles on its own.
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.
For each term, give its limit and say what the $n$th-term test concludes.
| Limit | The test concludes | |
|---|---|---|
| $\dfrac{n}{n + 1}$ | ||
| $\dfrac{1}{n^2}$ | ||
| $\dfrac{2n + 1}{n + 4}$ |
You can apply the $n$th-term test and state its limits. Without looking: what does the test conclude when the terms tend to zero, and which series is the reason it cannot conclude anything else?
9. Your turn: $\displaystyle\sum \cos\left(\frac{1}{n}\right)$, step 3