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Measuring a series of positive terms against a $p$-series or a geometric one: smaller than convergent and larger than divergent are the two combinations that settle anything, and limit comparison is what to reach for when the inequality will not come out the right 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 choose a yardstick series by keeping the dominant part of a term, establish the inequality between the two, and say which of the four direction-and-verdict combinations you are in — including recognising the two that are true and prove nothing. You can also apply limit comparison, checking that the ratio's limit is finite and positive before transferring the verdict.
You can settle a $p$-series and a geometric series on sight. Those two families are the yardsticks for everything else: almost any series of positive terms can be measured against one of them, and comparison is how the measurement is turned into a verdict.
Direct comparison: an inequality $a_n \le b_n$ or $a_n \ge b_n$ between the terms of two series of positive terms.
Limit comparison: the ratio $\lim \dfrac{a_n}{b_n}$, used when it is finite and positive.
Yardstick series: a $p$-series or a geometric series, whose verdict is already known.
Settles: the comparison, together with the yardstick's verdict, forces a verdict for the series in hand.
Both comparison tests require positive terms. With that in place, direct comparison says the obvious thing:
The other two combinations say nothing. Smaller than a divergent series, or larger than a convergent one, is compatible with either answer, and a true inequality in the wrong direction is the commonest way a comparison argument turns out to be empty. Every convergence test is a theorem with conditions attached, and a test quoted where its conditions fail has not been applied — it has been guessed with. Naming the condition you checked is part of the answer, and on an examination it is part of the marks.
Choosing the comparison is the part that takes judgement. Throw away everything that stops mattering for large $n$: $\dfrac{1}{n^2+1}$ behaves like $\dfrac{1}{n^2}$, and $\dfrac{3n+1}{n^3-2}$ behaves like $\dfrac{3}{n^2}$.
Limit comparison removes the need to get the inequality right. If $a_n$ and $b_n$ are positive and
$$\lim_{n\to\infty}\frac{a_n}{b_n} = L \quad\text{with } 0 < L < \infty,$$
then $\sum a_n$ and $\sum b_n$ do the same thing. This is what to reach for when the obvious comparison points the wrong way — $\dfrac{1}{\sqrt{n}-1}$ is bigger than $\dfrac{1}{\sqrt{n}}$, which is fine here, but $\dfrac{1}{n^2-n}$ against $\dfrac{1}{n^2}$ is bigger and the comparison would be useless, while the ratio tends to $1$ and settles it at once.
Another way: picture
Picture two stacks of coins built term by term. If yours is never taller than a stack known to stop below the ceiling, yours stops too. If yours is never shorter than a stack known to pass the ceiling, yours passes too. Knowing your stack is shorter than one that goes through the roof tells you nothing at all about where yours stops.
Another way: steps
| Your terms are | The yardstick | Conclusion |
|---|---|---|
| smaller | converges | converges |
| larger | diverges | diverges |
| smaller | diverges | nothing |
| larger | converges | nothing |
Half this table is useful and the other half is where marks are lost. The rule of thumb: a comparison only helps when it pushes your series toward the verdict the yardstick already has — under a finite ceiling, or over an infinite floor.
Getting the inequality direction backwards. A bigger denominator makes a smaller fraction, so $\dfrac{1}{n^2+1} < \dfrac{1}{n^2}$. Write the inequality out rather than guessing it from the shape.
Using a true comparison in a useless direction. The two dead combinations produce arguments that look complete and establish nothing. Check which of the four rows you are in before writing a conclusion.
Forgetting that limit comparison needs a limit that is finite and positive. A ratio tending to $0$ or to infinity gives only a one-way conclusion, and at this level the safe move is to choose a different yardstick so that the limit comes out in between.
$\displaystyle\sum \frac{1}{2^n + n}$: all terms are positive, and the yardstick is $\sum \dfrac{1}{2^n}$.
Keep the dominant part.
$2^n + n > 2^n$, so $\dfrac{1}{2^n+n} < \dfrac{1}{2^n}$: our terms are smaller.
Bigger denominator, smaller fraction.
The yardstick is geometric with $r = \tfrac12$ and converges. Smaller than convergent, so our series converges.
$\displaystyle\sum \frac{1}{n^2 - n}$ (from $n = 2$): the natural yardstick is $\sum \dfrac{1}{n^2}$, which converges.
But $n^2 - n < n^2$, so our terms are larger — larger than convergent, which settles nothing.
The dead combination.
Limit comparison instead: $\dfrac{1/(n^2-n)}{1/n^2} = \dfrac{n^2}{n^2-n} \to 1$, finite and positive, so the series converges after all.
For large $n$ the term behaves like $\dfrac{3n}{n^3} = \dfrac{3}{n^2}$, so the yardstick is $\sum \dfrac{1}{n^2}$.
Keep the dominant parts, top and bottom.
The ratio is $\dfrac{(3n+1)n^2}{n^3+2} \to 3$, which is finite and positive.
Limit comparison, so no inequality is needed.
$\sum \dfrac{1}{n^2}$ converges, so this series converges too.
Match each series to the simpler series it is naturally compared with.
| $\sum \dfrac{1}{n^2}$ | $\sum \dfrac{1}{2^n}$ | $\sum \dfrac{1}{\sqrt{n}}$ | $\sum \dfrac{1}{n}$ | |
|---|---|---|---|---|
| $\sum \dfrac{1}{n^2 + 1}$ | ||||
| $\sum \dfrac{1}{2^n + n}$ | ||||
| $\sum \dfrac{1}{\sqrt{n} - 1}$ | ||||
| $\sum \dfrac{\ln n}{n}$ |
For each series, say whether its terms are smaller or larger than the comparison's, and what that settles.
| Its terms are | So the series | |
|---|---|---|
| $\dfrac{1}{n^2 + 6}$ against $\dfrac{1}{n^2}$ | ||
| $\dfrac{n + 6}{n^2}$ against $\dfrac{1}{n}$ | ||
| $\dfrac{1}{2^n + 6}$ against $\dfrac{1}{2^n}$ |
The terms of $\displaystyle\sum \dfrac{1}{n^{5}}$ are smaller than those of $\displaystyle\sum \dfrac{1}{n}$, which diverges. What does that comparison establish?
Put the steps of settling $\displaystyle\sum \dfrac{1}{n^2 + 2}$ by comparison into the order you do them.
Number the steps in order (write the number in the box):
To settle $\displaystyle\sum \dfrac{6n + 4}{7n^2 + 8}$ by limit comparison with $\displaystyle\sum \dfrac1n$, find $\displaystyle\lim_{n\to\infty} \dfrac{a_n}{b_n}$.
Answer:
A series of positive terms is being tested. Select every statement below that settles it 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.
The terms of $\displaystyle\sum \dfrac{1}{n^{5}}$ are smaller than those of $\displaystyle\sum \dfrac{1}{n}$, which diverges. What does that comparison establish?
You can settle a series by comparison and say when a comparison is useless. Without looking: which two combinations of direction and verdict prove something, and what does limit comparison need of its limit?
9. Your turn: $\displaystyle\sum \frac{3n + 1}{n^3 + 2}$, step 3