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Both find a geometric series hiding inside the terms, both conclude absolute convergence, and both are blind at exactly the value one.
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 form and simplify a ratio or an $n$th root before taking its limit, choose the ratio test for a factorial and the root test for an $n$th power, compare the limit with one and recognise that the value one is a third outcome rather than a verdict, say why every $p$-series sits at that value and why that forces the tests to be silent there, report the convergence obtained as absolute, and read a ratio test applied to a series containing a variable as a condition on that variable.
Comparison needs a yardstick, and the yardsticks so far have been $p$-series and geometric series. That handles anything built from powers of $n$.
It does not handle $\sum \dfrac{2^n}{n!}$. There is no power of $n$ to compare a factorial with — $n!$ outgrows every power and every exponential, and no $p$-series is the right size. What is needed is a test that looks at how each term compares with the previous one, rather than with a fixed yardstick, and that is what the ratio test does.
The ratio is $\left|\dfrac{a_{n+1}}{a_n}\right|$ — how much bigger each term is than the last. The root is $\sqrt[n]{|a_n|}$ — the number that, raised to the $n$th power, gives the term.
A test is inconclusive when its hypotheses hold and its conclusion does not fire. Here that happens at exactly the value $1$, for both tests.
Both tests conclude absolute convergence — convergence of $\sum|a_n|$ — which is stronger than convergence and is why they may be applied to series with negative or alternating terms without any extra argument.
Ratio test. Let $L = \displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$. Then
Root test. Let $L = \displaystyle\lim_{n\to\infty}\sqrt[n]{|a_n|}$, with the same three cases.
Why they work. If $L < 1$, pick $r$ with $L < r < 1$. Eventually $|a_{n+1}| \le r|a_n|$, so from some point on the terms are dominated by a geometric series with ratio $r < 1$, which converges. Both tests are a comparison with a geometric series that the terms themselves reveal. If $L > 1$ the terms eventually grow, so they cannot tend to zero, and the divergence test ends it.
Why $L = 1$ is genuinely empty. Every $p$-series has $L = 1$ under both tests, and $p$-series include convergent and divergent cases. A test that concluded anything at $L = 1$ would therefore be false. This is the most important thing on the page.
Which to use. A factorial cancels cleanly in a ratio. An $n$th power comes apart cleanly under a root. When both apply they give the same answer: if $\left|\frac{a_{n+1}}{a_n}\right| \to L$ then $\sqrt[n]{|a_n|} \to L$ as well.
With a variable in the terms the ratio produces a condition on $|x|$ rather than a number, and that condition is a radius of convergence — the subject of unit 5.
Another way: picture
A bar chart of the terms. The ratio test measures the height of each bar against the one before it: if that ratio settles below one, the bars are eventually shrinking at least as fast as a geometric series and their total is finite. The root test measures each bar against $r^n$ directly. Both are asking the same question — is there a geometric series underneath this? — from two different directions.
Another way: steps
The point of both tests is that something cancels, and which one cancels depends on the shape.
Factorials cancel in ratios. $\dfrac{(n+1)!}{n!} = n+1$ — the whole factorial collapses to a single factor, because consecutive factorials differ by one multiplication. Under a root, $\sqrt[n]{n!}$ is a genuinely awkward object (it grows like $n/e$, by Stirling's formula), so the root test on a factorial is a much harder calculation for the same answer.
$n$th powers come apart under roots. $\sqrt[n]{\left(\frac{n}{2n+1}\right)^n} = \dfrac{n}{2n+1}$, and the exponent has vanished completely. Under a ratio, the same term gives $\dfrac{(n+1)^{n+1}(2n+1)^n}{n^n(2n+3)^{n+1}}$, which is correct and unpleasant.
Exponentials are fine either way. $\dfrac{2^{n+1}}{2^n} = 2$ and $\sqrt[n]{2^n} = 2$, so an exponential is equally easy under both.
Powers of $n$ are fine under neither. $\dfrac{(n+1)^p}{n^p} \to 1$ and $\sqrt[n]{n^p} \to 1$. Both tests are blind here, and that is what the $p$-series test is for.
So the choice is made by the term's shape, and the two tests together cover exactly what comparison cannot.
Reading $L = 1$ as a verdict. It is silence. Both readings — convergent and divergent — are wrong, and each is wrong for half the $p$-series.
Comparing before taking the limit. For $\sum \frac{5^n}{n!}$ the ratio $\frac{5}{n+1}$ exceeds $1$ for the first four terms. The test is about the limit, and early terms are irrelevant.
Forgetting the absolute values. The tests are stated for $|a_{n+1}/a_n|$. Without them a negative ratio can be misread.
Cancelling wrongly in a factorial. $\dfrac{(2n+2)!}{(2n)!} = (2n+2)(2n+1)$, not $2n+2$. Write out the two products if in doubt.
Using the ratio test on a $p$-series. It cannot work, and time spent on it is time lost.
Reporting plain convergence when the test gives absolute. The stronger statement is free and is worth having, because it is what permits rearrangement in lesson 23.
Assuming the limit exists. For some series the ratio oscillates and has no limit; the root test, stated with a $\limsup$, still applies and the ratio test does not. That is the sense in which the root test is strictly stronger.
Three outcomes is one more than most tests in this course have, and the third gets collapsed into one of the others almost universally. A learner who finds $L = 1$ and writes "so it diverges" will be right about half the time, which is the worst possible feedback.
The reason $L = 1$ must be empty is worth seeing rather than accepting. Every $p$-series has ratio limit $\left(\frac{n}{n+1}\right)^p \to 1$ and root limit $\left(n^{-p}\right)^{1/n} \to 1$ — for every $p$. And $p$-series include $\sum \frac{1}{n}$, which diverges, and $\sum \frac{1}{n^2}$, which converges. If either test concluded anything at $L = 1$, it would be concluding it for both, and it would be false.
So the value $1$ is not a near miss or a borderline that more care would resolve. It is the exact place where these tests are blind, and it is blind there because the family of series living at $L = 1$ is genuinely diverse. A ratio limit of $1$ says the terms are shrinking more slowly than geometrically — and within that range live all the interesting cases.
When you get $1$, the correct response is to stop and change tools: $p$-series, comparison, or the integral test. Those are precisely the tests that see what ratio and root cannot.
$\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^{n}}$: the ratio is $\dfrac{(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^{n}}{n!} = \dfrac{n^{n}}{(n+1)^{n}} = \left(\dfrac{n}{n+1}\right)^{n}$.
The factorial collapses to one factor, and so does the extra power.
$\left(\dfrac{n}{n+1}\right)^{n} = \left(1 + \dfrac1n\right)^{-n} \to \dfrac1e$, using the limit that defines $e$.
A sequence limit from lesson 17.
$\dfrac1e \approx 0.368 < 1$, so the series converges absolutely. Note how narrowly: if the limit had been $1$ the test would have said nothing, and $n!/n^n$ is not far from that boundary.
$\displaystyle\sum_{n=1}^{\infty}\left(\frac{3n + 1}{4n + 5}\right)^{n}$: the whole term is raised to the $n$, which is the root test's signal.
Shape first.
$\sqrt[n]{|a_n|} = \dfrac{3n+1}{4n+5} \to \dfrac34$, and the exponent has gone entirely.
One line, where a ratio would be several.
$\tfrac34 < 1$, so the series converges absolutely. Changing the $4$ to a $3$ would give a limit of $1$ and the test would say nothing — and indeed $\left(\frac{3n+1}{3n+5}\right)^n \to e^{-4/3} \ne 0$, so that series diverges by the divergence test instead.
Factorials, so the ratio test. $\dfrac{a_{n+1}}{a_n} = \dfrac{(2n+2)!}{((n+1)!)^{2}4^{n+1}}\cdot\dfrac{(n!)^{2}4^{n}}{(2n)!}$.
Write both out before cancelling.
$\dfrac{(2n+2)!}{(2n)!} = (2n+2)(2n+1)$ and $\dfrac{(n!)^{2}}{((n+1)!)^{2}} = \dfrac{1}{(n+1)^{2}}$, so the ratio is $\dfrac{(2n+2)(2n+1)}{4(n+1)^{2}}$.
Two factors from the top factorial, not one.
That simplifies to $\dfrac{2n+1}{2n+2} \to 1$. The ratio test is inconclusive — and the series does in fact diverge, which a finer argument (the terms behave like $1/\sqrt{\pi n}$) establishes.
Find $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ for $a_n = \dfrac{n^{4}}{2^{\,n}}$.
Answer:
For $\displaystyle\sum\frac{1}{n^{5}}$ the ratio limit is exactly $1$. What does the ratio test conclude?
Put the steps of applying the ratio test to $\displaystyle\sum\frac{8^{\,n}}{n!}$ into the order you do them.
Number the steps in order (write the number in the box):
Give the ratio limit and the verdict for each series.
| Ratio limit | What the ratio test says | |
|---|---|---|
| $\sum \dfrac{3^{\,n}}{n!}$ | ||
| $\sum \dfrac{n!}{3^{\,n}}$ | ||
| $\sum \dfrac{1}{n^{3}}$ |
Match each series to the test its term's shape calls for.
| The ratio test | The root test | Neither — use the $p$-series test | |
|---|---|---|---|
| $\sum \dfrac{3^{\,n}}{n!}$ | |||
| $\sum \left(\dfrac{n}{3n + 1}\right)^{n}$ | |||
| $\sum \dfrac{1}{n^{3}}$ | |||
| $\sum \dfrac{n^{3}}{4^{\,n}}$ |
For $\displaystyle\sum_{n=0}^{\infty}5^{\,n}x^{n}$, the ratio test gives a condition on $|x|$ rather than a verdict. For which $|x|$ is the ratio limit exactly $1$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ for $a_n = \dfrac{n^{2}}{3^{\,n}}$.
Answer:
You can settle a series with a factorial or an $n$th power in it, and stop when the test goes silent. Say in your own words why a ratio limit of exactly one cannot possibly decide anything. Next: series whose signs alternate, where a whole new kind of convergence appears.
10. Your turn: does $\displaystyle\sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^{2}4^{n}}$ converge?, step 3
Inconclusive is an honest outcome.