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Trapping the partial sums against a $p$-series or a geometric one, in whichever of the two useful directions the verdict requires.
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 choose a yardstick series by keeping the dominant terms, decide its verdict before writing any inequality, recognise which two of the four possible pairings settle nothing, rescue an inequality that points the wrong way by scaling the yardstick with a constant, apply limit comparison and state what a finite non-zero limit does and does not tell you, handle the edge cases where the ratio tends to zero or infinity, and check that the terms are non-negative before applying either test.
Unit 3 settled improper integrals by comparison: trapped below something convergent, or forced above something divergent. The proof was that $\int_a^b f$ increases with $b$ and is bounded, so it converges.
Series work the same way with partial sums in place of truncated integrals. If $a_n \ge 0$ then $s_N$ increases with $N$, so the series converges exactly when the partial sums are bounded — and a comparison is a way of bounding them. Every test in this lesson is that sentence applied.
A yardstick is a series whose verdict you already know: a $p$-series, a geometric series, or something already settled.
The dominant term of an expression in $n$ is the part that decides its size for large $n$. Keeping the dominant terms above and below and cancelling gives the yardstick.
Eventually means "for all $n$ beyond some point". Every test in this lesson only needs its inequality to hold eventually, because finitely many terms are a finite sum and cannot change a verdict.
Throughout, all terms are non-negative; this is essential and is checked, not assumed.
Comparison test. If $0 \le a_n \le b_n$ eventually, then
And nothing else. Below a divergent series settles nothing; above a convergent one settles nothing. Two of the four pairings are useless, and they are the two that get used.
Limit comparison test. If $a_n, b_n > 0$ and $$\lim_{n\to\infty}\frac{a_n}{b_n} = L \quad\text{with } 0 < L < \infty,$$ then $\sum a_n$ and $\sum b_n$ converge or diverge together. This is usually the easier tool, because no direction has to be chosen and no inequality has to be argued.
The edge cases. $L = 0$ gives only $\sum b_n$ converges $\implies$ $\sum a_n$ converges; $L = \infty$ gives only $\sum b_n$ diverges $\implies$ $\sum a_n$ diverges. Quoting the two-way equivalence in those cases is a real error.
Choosing the yardstick. Keep the dominant power on top and bottom and cancel. $\dfrac{3n + 5}{n^3 - 2}$ behaves like $\dfrac{3n}{n^3} = \dfrac{3}{n^2}$, and constants do not matter, so compare with $\sum n^{-2}$.
Scaling is free. $\sum c\,b_n$ has the same verdict as $\sum b_n$ for any $c > 0$. So an inequality that fails by a constant factor can be rescued by putting the constant in.
Another way: picture
Two stacks of blocks, one block per term, side by side. If the taller stack is finite, the shorter one cannot be infinite. If the shorter stack is infinite, the taller one cannot be finite. Those are the two useful statements, and the picture also shows why the other two say nothing: a short stack beside an infinite one, or a tall stack beside a finite one, is unconstrained.
Another way: steps
Direct comparison requires an inequality, and the inequality has to point the way you want. That is often awkward.
Take $\sum \dfrac{1}{n^2 - 3}$. It behaves like $\sum n^{-2}$, which converges, so you want the terms below a convergent series. But $\dfrac{1}{n^2 - 3} > \dfrac{1}{n^2}$ for every $n$ — a smaller denominator makes a bigger fraction — so the obvious comparison is useless. It can be rescued: for $n \ge 3$, $n^2 - 3 \ge \tfrac12 n^2$, so the terms are below $\dfrac{2}{n^2}$, and $\sum \dfrac{2}{n^2}$ converges. Two extra lines and a constant chosen by hand.
Limit comparison does it without any of that: $\dfrac{1/(n^2-3)}{1/n^2} = \dfrac{n^2}{n^2 - 3} \to 1$, finite and non-zero, so the two series agree. One line, no inequality, no direction to get right.
The general rule: use limit comparison when the term is a ratio of polynomial-like expressions, which is most of the time. Reach for a direct comparison when the term has something a limit cannot handle cleanly — a $\sin n$ or a $\ln n$ that needs bounding rather than evaluating — or when a bound on the sum is wanted as well as a verdict.
Comparing series with negative terms. The test is stated for non-negative terms and the proof needs them: without positivity the partial sums need not increase, and boundedness proves nothing.
Using a useless direction. Below divergent, or above convergent, is not a weaker argument — it is not an argument. This is where most of the marks go.
Choosing a yardstick that is not simpler. Comparing with something you have not settled leaves you with two unsolved problems.
Ignoring the edge cases of limit comparison. $L = 0$ and $L = \infty$ give one-directional information only.
Insisting the inequality hold for every $n$. Eventually is enough.
Reporting the ratio's limit as a sum. It is neither series' sum.
Forgetting to state the yardstick's verdict. "It behaves like $n^{-1/2}$" is half an argument; the other half is that $\sum n^{-1/2}$ diverges.
Most attempts at these problems get the yardstick right and then stop. "$\dfrac{1}{n^2+1}$ behaves like $\dfrac{1}{n^2}$, which converges, so it converges" is the shape of nearly every answer, and it is missing the only part that could be wrong.
The part that is missing is the direction. Behaving alike is not a theorem; the theorems are below convergent and above divergent, plus limit comparison, and each requires something to be established. Here the inequality happens to be easy — $\dfrac{1}{n^2 + 1} < \dfrac{1}{n^2}$ — and a learner who never writes it down will not notice that in $\dfrac{1}{n^2 - 1}$ it points the other way and the same sentence is then unjustified.
There are two honest ways to finish. Either write the inequality, in the direction the yardstick's verdict requires, scaling by a constant if needed; or take the limit of the ratio and quote limit comparison. Both take one line. Neither is optional.
The habit to build: verdict of the yardstick first, then the direction you need, then the argument. Choosing the yardstick is the easy part, and it is the part everyone does.
$\displaystyle\sum_{n=1}^{\infty}\frac{2 + \sin n}{n^{2}}$: the numerator wobbles and has no limit, so limit comparison with $n^{-2}$ is awkward.
A term a limit cannot evaluate.
But $\sin n$ is always between $-1$ and $1$, so $\dfrac{2 + \sin n}{n^{2}} \le \dfrac{3}{n^{2}}$, and the terms are positive since $2 + \sin n \ge 1 > 0$.
Bound it rather than evaluating it.
$\sum \dfrac{3}{n^{2}}$ converges, and our terms are below it, so the series converges. The constant $3$ was free, which is what made a direct comparison the right tool here.
$\displaystyle\sum_{n=1}^{\infty}\sin\frac{1}{n}$: the terms are positive for $n \ge 1$ and tend to zero, so the divergence test is silent.
Check positivity and the terms first.
For small $x$, $\sin x \approx x$. Take $b_n = \dfrac1n$: $\dfrac{\sin(1/n)}{1/n} \to 1$ as $n \to \infty$, since $\dfrac{\sin t}{t} \to 1$ as $t \to 0$.
The standard trigonometric limit from Calculus I.
Finite and non-zero, so the two series agree — and $\sum \frac1n$ diverges. So $\sum \sin\frac1n$ diverges, although its terms are smaller than $\frac1n$ at every step.
Terms are positive. Dominant behaviour: $\dfrac{n}{n^{3}} = \dfrac{1}{n^{2}}$, so the yardstick is $\sum n^{-2}$, which converges.
Yardstick and its verdict first.
Limit comparison: $\dfrac{(n+4)/(n^{3}+2n)}{1/n^{2}} = \dfrac{n^{3} + 4n^{2}}{n^{3} + 2n} \to 1$.
Finite and non-zero.
So the series converges. A direct comparison would also work here, since $\dfrac{n+4}{n^3 + 2n} \le \dfrac{5n}{n^3} = \dfrac{5}{n^2}$ for $n \ge 1$ — but it needed a constant chosen by hand.
Read the claim with sums in place of integrals: $0 \le g \le f$ on $[1, \infty)$ and $\int_1^\infty g$ diverges. Does it settle whether $\sum a_n$ converges, where $a_n = f(n)$ and $b_n = g(n)$?
Match each series to the $p$-series it should be compared with.
| $\sum \dfrac{1}{n^{2}}$ | $\sum \dfrac{1}{\sqrt{n}}$ | $\sum \dfrac{1}{n^{3}}$ | |
|---|---|---|---|
| $\sum \dfrac{1}{n^{2} + 6}$ | |||
| $\sum \dfrac{n}{n^{3} + 6}$ | |||
| $\sum \dfrac{1}{\sqrt{n} + 6}$ | |||
| $\sum \dfrac{n^{2} + 6}{n^{5}}$ |
To test $\displaystyle\sum_{n=1}^{\infty}\frac{5n + 1}{8n^{3} + n}$ by limit comparison with $\displaystyle\sum\frac{1}{n^{2}}$, find $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}$.
Answer:
For each series, give the exponent of the $p$-series it behaves like, and the verdict.
| Exponent | Verdict | |
|---|---|---|
| $\sum \dfrac{1}{n^{3} + 2}$ | ||
| $\sum \dfrac{n}{n^{2} + 2}$ | ||
| $\sum \dfrac{1}{n\sqrt{n} + 2}$ |
Put the steps of settling $\displaystyle\sum\frac{1}{n^{2} - 7}$ by comparison into the order you do them.
Number the steps in order (write the number in the box):
To prove $\displaystyle\sum_{n > 4}\frac{1}{n^{2} - 4 n}$ converges you need $\dfrac{1}{n^{2} - 4n} \le \dfrac{C}{n^{2}}$ with $C = 2$. From which $n$ onwards does that hold?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Read the claim with sums in place of integrals: $0 \le f \le g$ on $[1, \infty)$ and $\int_1^\infty g$ diverges. Does it settle whether $\sum a_n$ converges, where $a_n = f(n)$ and $b_n = g(n)$?
You can settle a positive series by comparison and name the direction your argument runs in. Say in your own words why $\sum \sin(1/n)$ diverges although every term is smaller than $1/n$. Next: the tests for factorials and $n$th powers, where comparison has nothing to offer.
10. Your turn: does $\displaystyle\sum_{n=1}^{\infty}\frac{n + 4}{n^{3} + 2n}$ converge?, step 3