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The limit of $\left|\dfrac{a_{n+1}}{a_n}\right|$: below one the series converges absolutely, above one it diverges, and at exactly one the test is silent — necessarily so, since $\sum n^{-1}$ and $\sum n^{-2}$ both produce it.
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 form the ratio of consecutive terms, cancel factorials and powers correctly, take its limit and read the verdict off a comparison with one. You can recognise in advance the series the test is built for — factorials and constants raised to the $n$ — and the ones on which it is guaranteed to return one, and you can say why a limit of one has to leave the question open.
You can sum a geometric series and you know it converges exactly when the ratio between consecutive terms has size below one. The ratio test asks the same question of a series that is not geometric: does the ratio between consecutive terms eventually settle below one, and if so, the series is beaten by a geometric one.
The ratio test: compute $L = \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$. If $L < 1$ the series converges absolutely; if $L > 1$ it diverges; if $L = 1$ the test is inconclusive.
Absolutely convergent: $\sum |a_n|$ converges, which is stronger than plain convergence.
Inconclusive: the test returns nothing and another test is needed.
If the ratio of consecutive terms settles down to some number $L$, then far out the series behaves like a geometric series with ratio $L$. Geometric series converge exactly when the ratio has size below one, and comparison does the rest:
$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \begin{cases} < 1 & \text{converges absolutely} \\ > 1 & \text{diverges} \\ = 1 & \text{no conclusion} \end{cases}$$
Note the absolute values. The ratio is taken of sizes, so the test never cares about signs, and what it delivers when $L < 1$ is absolute convergence — the strongest form there is.
What the test is for. Factorials and constants raised to the $n$, because those are exactly the things whose ratio simplifies. $\dfrac{a_{n+1}}{a_n}$ turns $(n+1)!$ over $n!$ into $n+1$, and $c^{n+1}$ over $c^{n}$ into $c$. A page of algebra becomes one line.
What it is useless for. Anything that is a ratio of powers of $n$. $\left(\dfrac{n}{n+1}\right)^{k} \to 1$ for every $k$, so the answer is $1$ and the test is silent — every time, predictably. On a $p$-series, reach for the $p$-rule instead and do not spend the two minutes.
The silence at one is forced. $\sum \tfrac1n$ diverges, $\sum \tfrac1{n^2}$ converges, and both give $L = 1$. Any rule that produced a verdict from $L = 1$ would have to give these two the same one.
Another way: picture
Picture the terms as a sequence of shrinking steps and ask by what factor each step is shorter than the last. If that factor settles at nine tenths, the steps are eventually a geometric decay and the total distance is finite. If it settles at one, the steps are shrinking by less and less each time — and whether the journey ends depends on details the factor alone cannot see.
Another way: steps
| The term contains | The ratio collapses to | Worth using? |
|---|---|---|
| A factorial | something over $n+1$, so $0$ | always |
| A constant to the $n$ | that constant | always |
| Both | the constant | always |
| Powers of $n$ only | $1$ | never |
The last row can be predicted before any algebra is done, which makes it the single most useful line in the table: it tells you not to start.
Reading $L = 1$ as a verdict. It is not the series converges, nor the series diverges. It is this test has nothing to say, and writing anything else is a claim the working does not support.
Inverting the ratio. It is the next term over the current one. Turning it upside down turns every verdict into its opposite and every answer will look internally consistent.
Cancelling factorials carelessly. $\dfrac{(n+1)!}{n!} = n+1$, not $n!$ and not $1$. Write $(n+1)! = (n+1)\cdot n!$ explicitly the first few times rather than doing it in your head.
$\displaystyle\sum \frac{5^{n}}{n!}$: the ratio is $\dfrac{5^{n+1}}{(n+1)!}\cdot\dfrac{n!}{5^{n}}$.
Write it out before cancelling.
The powers leave $5$ and the factorials leave $n+1$ underneath, so the ratio is $\dfrac{5}{n+1}$.
$(n+1)! = (n+1)\cdot n!$
That tends to $0 < 1$, so the series converges absolutely — and quickly, since the ratio is eventually tiny.
$\displaystyle\sum \frac{n}{n^2 + 1}$: the ratio is $\dfrac{n+1}{(n+1)^2+1}\cdot\dfrac{n^2+1}{n}$.
Powers of $n$ only.
Every factor is a ratio of polynomials of equal degree, so the whole thing tends to $1$.
Predictable in advance.
The test concludes nothing. Limit comparison with $\sum \tfrac1n$ gives a ratio of $1$ and settles it: the series diverges.
The ratio is $\dfrac{(n+1)!}{10^{n+1}}\cdot\dfrac{10^{n}}{n!}$.
Next term over current.
The factorials leave $n+1$ on top and the powers leave $10$ underneath, so the ratio is $\dfrac{n+1}{10}$.
This time the factorial is upstairs.
That grows without limit, so $L > 1$ and the series diverges. A factorial in the numerator beats any exponential, however large its base.
Put the steps of applying the ratio test to $\displaystyle\sum \dfrac{4^{n}}{n!}$ into the order you do them.
Number the steps in order (write the number in the box):
For each series, give the limiting ratio and say what the ratio test concludes.
| Limiting ratio | The test concludes | |
|---|---|---|
| $\sum \dfrac{2^{n}}{n!}$ | ||
| $\sum \dfrac{2^{n}}{n^2}$ | ||
| $\sum \dfrac{1}{n}$ |
For $\sum \dfrac{n}{2^n}$, find the limit of $\left|\dfrac{a_{n+1}}{a_n}\right|$ as $n$ grows.
Answer:
The ratio test is applied to $\sum \dfrac{n}{2^n}$, and the limiting ratio comes out as $\dfrac{1}{2}$. What does the test conclude?
Match each series to what its limiting ratio turns out to be.
| $L = 0$, so it converges | $L = 3$, so it diverges | $L = 1$, so the test is silent | $L = \tfrac12$, so it converges | |
|---|---|---|---|---|
| $\sum \dfrac{2^n}{n!}$ | ||||
| $\sum \dfrac{3^n}{n^2}$ | ||||
| $\sum \dfrac{1}{n^2}$ | ||||
| $\sum \dfrac{n}{2^n}$ |
The ratio test is applied to $\sum \dfrac{1}{n}$ and to $\sum \dfrac{1}{n^{2}}$. Select every statement below that is true.
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.
Put the steps of applying the ratio test to $\displaystyle\sum \dfrac{6^{n}}{n!}$ into the order you do them.
Number the steps in order (write the number in the box):
You can apply the ratio test and say where it is useless. Without looking: what are the three cases for the limiting ratio, and which pair of series forces the test to stay silent at one?
9. Your turn: $\displaystyle\sum \frac{n!}{10^{n}}$, step 3