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Calculus BC · Unit 10 · 10.6
AP Calculus BC · Infinite Sequences and Series
Comparison Tests for Convergence
Swap an awkward series for a familiar one — by inequality, or just by behaviour.
key idea$0\le a_n\le b_n:\ \ \sum b_n\ \text{converges}\Rightarrow\sum a_n\ \text{converges}$
The test
Applies to
series with non-negative terms only
Direct comparison
smaller than convergent $\Rightarrow$ converges; larger than divergent $\Rightarrow$ diverges
Limit comparison
if $\displaystyle\lim\frac{a_n}{b_n}=L$ with $0\lt L\lt\infty$, both series share a verdict
Useless directions
smaller than divergent, or larger than convergent — these say nothing
Usual partner
a $p$-series or a geometric series
Comparison replaces an awkward series with a familiar one. The direct form needs a genuine inequality, which is often fiddly to arrange; the limit form needs only that the two series behave alike for large $n$, which is far easier and almost always the better choice.
| You know | Your series is | Conclusion |
|---|
| $\sum b_n$ converges | smaller: $a_n\le b_n$ | converges |
| $\sum b_n$ diverges | larger: $a_n\ge b_n$ | diverges |
| $\sum b_n$ converges | larger | no conclusion |
| $\sum b_n$ diverges | smaller | no conclusion |
| $\displaystyle\lim\frac{a_n}{b_n}=L\in(0,\infty)$ | either | same verdict as $\sum b_n$ |
Rows 3 and 4 are where direct comparison fails, and they are exactly why the limit comparison test exists: it needs no inequality at all, only a finite positive ratio.
Direct comparisonUses a term-by-term inequality. Requires you to actually prove $a_n\le b_n$ for all large $n$.
Limit comparisonUses $\displaystyle\lim\frac{a_n}{b_n}$. A finite positive limit means the two series rise or fall together.
Comparison partnerThe familiar series you compare against — nearly always a $p$-series or a geometric series.
Dominant termThe part of $a_n$ that decides its size for large $n$. Keeping only the leading terms of numerator and denominator picks the partner.
When to reach for it
| If $a_n$ looks like… | Compare with | Which form |
|---|
| a ratio of polynomials | $\dfrac{1}{n^{\,\deg\text{denom}-\deg\text{num}}}$ | limit |
| $\dfrac{1}{n^{2}+n}$ | $\dfrac{1}{n^{2}}$ | either — the inequality is easy |
| $\dfrac{\sin^{2}n}{n^{2}}$ | $\dfrac{1}{n^{2}}$ | direct, since $\sin^{2}n\le1$ |
| $\dfrac{1}{2^{n}+n}$ | $\dfrac{1}{2^{n}}$ | direct |
| $\dfrac{\sqrt{n}+1}{n^{2}-3}$ | $\dfrac{1}{n^{3/2}}$ | limit |
Choose the partner by keeping only the dominant term of the numerator and of the denominator. If the resulting inequality is awkward to prove, switch to the limit form — it needs no inequality.
Worked examples
EX 1
Direct comparison
a bounded numeratorDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{\sin^{2}n}{n^{2}}$ converges.
1
Bound the numerator
$0\le\sin^{2}n\le1$ for every $n$.
2
Build the inequality
$0\le\dfrac{\sin^{2}n}{n^{2}}\le\dfrac{1}{n^{2}}$.
3
Identify the partner
$\displaystyle\sum\dfrac{1}{n^{2}}$ is a $p$-series with $p=2\gt 1$, so it converges.
4
Apply direct comparison
A non-negative series bounded above by a convergent one converges.
5
Conclude
The series converges. Limit comparison would fail here, since $\dfrac{a_n}{b_n}=\sin^{2}n$ has no limit.
EX 2
Limit comparison
polynomialsDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{2n+1}{n^{3}-4}$ converges.
1
Find the dominant behaviour
$\dfrac{2n}{n^{3}}=\dfrac{2}{n^{2}}$, so compare with $b_n=\dfrac{1}{n^{2}}$.
2
Form the ratio
$\dfrac{a_n}{b_n}=\dfrac{\left(2n+1\right)n^{2}}{n^{3}-4}=\dfrac{2n^{3}+n^{2}}{n^{3}-4}$.
3
Take the limit
$\displaystyle\lim_{n\to\infty}\dfrac{2n^{3}+n^{2}}{n^{3}-4}=2$.
4
Check the limit qualifies
$0\lt 2\lt\infty$, so the test applies and both series share a verdict.
5
Conclude
$\displaystyle\sum\dfrac{1}{n^{2}}$ converges, so the series converges.
EX 3
Comparison in the divergent direction
larger than harmonicDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}-1}$ converges, for $n\ge2$.
1
Choose the partner
For large $n$ the term behaves like $\dfrac{1}{\sqrt{n}}$, a $p$-series with $p=\dfrac12$, which diverges.
2
Set up the inequality
$\sqrt{n}-1\lt\sqrt{n}$, so $\dfrac{1}{\sqrt{n}-1}\gt\dfrac{1}{\sqrt{n}}$.
3
Check the direction
Our series is larger than a divergent one — a usable direction.
4
Apply direct comparison
Larger than divergent forces divergence.
5
Conclude
The series diverges. Note that larger-than-convergent would have told us nothing.
EX 4
When the ratio limit is zero
the extra clauseA student compares $a_n=\dfrac{\ln n}{n^{3}}$ with $b_n=\dfrac{1}{n^{2}}$ and finds $\displaystyle\lim\dfrac{a_n}{b_n}=0$. What can be concluded?
1
Compute the ratio
$\dfrac{a_n}{b_n}=\dfrac{\ln n}{n}$, whose limit is $0$.
2
Note that $L=0$ is outside the main clause
The standard statement requires $0\lt L\lt\infty$.
3
Use the one-sided version
If $L=0$ and $\sum b_n$ converges, then $\sum a_n$ converges as well — the terms are eventually much smaller.
4
Apply it
$\displaystyle\sum\dfrac{1}{n^{2}}$ converges, so the series converges.
5
Note the limitation
Had $\sum b_n$ diverged, $L=0$ would have given no information. The direction still matters.
What it cannot tell you
Two of the four directions are useless
Smaller than divergent, and larger than convergent, both give no conclusion. Check the direction before writing anything down.
Terms must be non-negative
Comparison tests assume $a_n\ge0$. For alternating series, compare $\lvert a_n\rvert$ and you are testing absolute convergence instead.
Limit comparison needs $0\lt L\lt\infty$
A limit of $0$ or $\infty$ only works in one direction, and it must be stated which. The safe move is to choose a partner giving a finite positive limit.
The partner must be one you know
Comparing against a series whose behaviour you have not established proves nothing. In practice that means a $p$-series or a geometric series.