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Calculus BC · Unit 10 · 10.9
AP Calculus BC · Infinite Sequences and Series
Determining Absolute or Conditional Convergence
Strip the signs and test that series first — a convergent answer there settles everything.
key idea$\text{Test }\sum\lvert a_n\rvert\ \text{first: converges}\Rightarrow\text{absolute; diverges}\Rightarrow\text{check }\sum a_n$
How to choose
By this point you own seven convergence tests, and the question is which to reach for. The answer starts with a single decision: does the series have negative terms? If it does, test the absolute-value series first — a convergent answer there settles everything at once.
| Term looks like… | Try | Because |
|---|
| $ar^{n}$ — constant ratio | geometric | gives a verdict and a sum |
| $n^{-p}$ | $p$-series | one glance at the exponent |
| a ratio of polynomials | limit comparison with a $p$-series | degrees give the partner |
| contains $n!$ or $r^{n}$ | ratio test | factorials and powers cancel in the ratio |
| contains $\ln n$ | integral test | logarithms integrate by substitution |
| contains $(-1)^{n}$ | absolute value first, then alternating | settles absolute versus conditional |
| $a_n\not\to0$ | $n$th-term test | one limit, immediate divergence |
Run the last row first every time — it is the cheapest check available. After that, match the structure of the term to the test built for it rather than trying tests in a fixed order.
The decision tree
FIRSTDo the terms fail to approach zero?Then the series diverges by the $n$th-term test, and no further work is needed. This check costs one limit and rules out a whole class of series.
SIGNSAre there negative terms?If every term is non-negative, skip ahead. If the signs alternate or vary, form $\sum\lvert a_n\rvert$ and test that series first.
ABSOLUTEDoes $\sum\lvert a_n\rvert$ converge?Then $\sum a_n$ converges absolutely — and absolute convergence implies ordinary convergence, so you are finished.
CONDITIONAL$\sum\lvert a_n\rvert$ diverges, but the terms alternate?Apply the alternating series test to $\sum a_n$. If it converges, the series converges conditionally. If the alternating test also fails, the series diverges.
STRUCTUREAll terms non-negative — what does $a_n$ contain?Factorials or $n$th powers → ratio test. A clean power of $n$ → $p$-series. A polynomial ratio → limit comparison. A logarithm → integral test.
STUCKNothing fits?Re-examine the dominant behaviour of $a_n$ for large $n$ and look again for a comparison partner. Almost every AP series reduces to a $p$-series or a geometric series in disguise.
Absolute$\sum\lvert a_n\rvert$ converges
Conditional$\sum a_n$ converges, $\sum\lvert a_n\rvert$ does not
Divergentneither converges
Worked examples
EX 1
Absolutely convergent
the absolute test succeedsClassify $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n^{2}}$.
1
Check the terms go to zero
$\dfrac{1}{n^{2}}\to0$, so the $n$th-term test gives nothing.
2
Form the absolute series
$\displaystyle\sum\dfrac{1}{n^{2}}$.
3
Test it
A $p$-series with $p=2\gt 1$, so it converges.
4
Draw the conclusion
Absolute convergence implies convergence, so nothing further is needed.
5
Classify
The series converges absolutely.
EX 2
Conditionally convergent
the absolute test failsClassify $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n}$.
1
Form the absolute series
$\displaystyle\sum\dfrac{1}{n}$, the harmonic series.
2
Test it
$p=1$, so it diverges. Absolute convergence is ruled out.
3
Return to the original
The terms alternate, so try the alternating series test.
4
Check its conditions
$b_n=\dfrac1n$ decreases and tends to $0$. ✓ The series converges.
5
Classify
Converges, but not absolutely — conditionally convergent.
EX 3
Divergent on the cheapest test
stop earlyClassify $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\dfrac{2n}{3n+1}$.
1
Take the limit of $\lvert a_n\rvert$
$\dfrac{2n}{3n+1}\to\dfrac23$.
2
Note that the terms do not vanish
$a_n$ oscillates near $\pm\dfrac23$, so $\displaystyle\lim a_n$ does not exist.
3
Apply the $n$th-term test
The series diverges.
4
Resist the wrong route
Testing $\sum\lvert a_n\rvert$ would also show divergence, but that alone would not settle $\sum a_n$.
5
Classify
Divergent. The first branch of the tree ended it in one line.
EX 4
Routing by structure
factorials call the ratio testClassify $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n}\,3^{n}}{n!}$.
1
Note the structure
A power and a factorial — the ratio test is built for this.
2
Apply it to the absolute values
$\dfrac{\lvert a_{n+1}\rvert}{\lvert a_n\rvert}=\dfrac{3^{n+1}}{(n+1)!}\cdot\dfrac{n!}{3^{n}}=\dfrac{3}{n+1}$.
3
Take the limit
$\displaystyle\lim_{n\to\infty}\dfrac{3}{n+1}=0\lt 1$.
4
Read the conclusion
$\sum\lvert a_n\rvert$ converges, so the original converges absolutely.
5
Note the efficiency
The ratio test tests absolute values by construction, so it answers both questions at once.
EX 5
A polynomial ratio with signs
comparison after strippingClassify $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n}\left(n+2\right)}{n^{2}+1}$.
1
Form the absolute series
$\displaystyle\sum\dfrac{n+2}{n^{2}+1}$, which behaves like $\dfrac{1}{n}$.
2
Limit-compare with $\dfrac1n$
$\displaystyle\lim\dfrac{\left(n+2\right)n}{n^{2}+1}=1$, finite and positive.
3
Read that verdict
$\sum\dfrac1n$ diverges, so $\sum\lvert a_n\rvert$ diverges.
4
Test the alternating series
$b_n=\dfrac{n+2}{n^{2}+1}\to0$ and is eventually decreasing, so $\sum a_n$ converges.
5
Classify
Conditionally convergent — the same two-stage route as the alternating harmonic series.
Watch out
Absolute convergence implies convergence
If $\sum\lvert a_n\rvert$ converges you are finished. There is no need to test the original series as well.
Divergence of $\sum\lvert a_n\rvert$ settles nothing
It rules out absolute convergence only. The original series may still converge conditionally, and that requires its own test.
Name the test you used
Free-response scoring rewards the citation. ‘Converges by the alternating series test, and $\sum\lvert a_n\rvert$ diverges by the $p$-series test with $p=1$’ is the complete answer.
Check the $n$th-term test first
It is one limit and it ends a whole class of problems immediately. Reaching for a comparison before checking it wastes minutes.