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Calculus BC · Unit 10 · 10.7
AP Calculus BC · Infinite Sequences and Series
The Alternating Series Test
Shrink steadily and reach zero, and the alternating signs do the rest.
key idea$\sum(-1)^{n}b_n\ \text{converges if}\ b_n\ \text{decreases and}\ b_n\to 0$
The test
Applies to
series whose terms alternate in sign: $\sum(-1)^{n}b_n$ with $b_n\gt 0$
Condition 1
$b_n$ is decreasing — $b_{n+1}\le b_n$ for all large $n$
Condition 2
$\displaystyle\lim_{n\to\infty}b_n=0$
Conclusion
the series converges. Both conditions are required
Cannot show
divergence. If $b_n\not\to0$ use the $n$th-term test instead
four requirements
- Strip the sign first: $b_n=\lvert a_n\rvert$, and check the two conditions on $b_n$.
- $b_n\to0$ alone is not enough — the terms must also shrink monotonically.
- Prove ‘decreasing’ by a ratio, a difference, or the derivative of the matching function.
- The test gives convergence only; it says nothing about whether that convergence is absolute.
Alternating signs let partial sums bracket the limit: each new term overshoots the target by less than the last. That is why the terms must shrink steadily as well as reach zero — without monotonicity, the bracketing argument collapses.
| Series | $b_n$ | Decreasing? | $b_n\to0$? | Verdict |
|---|
| $\displaystyle\sum\frac{(-1)^{n}}{n}$ | $\tfrac1n$ | yes | yes | converges |
| $\displaystyle\sum\frac{(-1)^{n}}{\sqrt{n}}$ | $n^{-1/2}$ | yes | yes | converges |
| $\displaystyle\sum(-1)^{n}\frac{n}{n+1}$ | $\tfrac{n}{n+1}$ | no | no — $\to1$ | diverges |
| $\displaystyle\sum\frac{(-1)^{n}}{n^{2}}$ | $n^{-2}$ | yes | yes | converges |
Row 1 is the alternating harmonic series — it converges, while $\sum\tfrac1n$ does not. The alternating signs are doing all the work, and that distinction is what conditional convergence is built on.
When to reach for it
| Sign of the term | First move | Then |
|---|
| $(-1)^{n}$ or $(-1)^{n+1}$ present | strip it to get $b_n$ | check the two conditions |
| $\cos(n\pi)$ present | rewrite as $(-1)^{n}$ | same as above |
| $b_n\not\to0$ | stop | the $n$th-term test gives divergence |
| $b_n\to0$ but not monotone | check further out | monotone eventually is enough |
| all terms positive | this test does not apply | use comparison, ratio or integral |
Row 4 matters more than it looks: the conditions only need to hold from some point onward. Finitely many misbehaving terms cannot change convergence.
Worked examples
EX 1
The alternating harmonic series
both conditionsDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}$ converges.
1
Strip the sign
$b_n=\dfrac1n\gt 0$.
2
Check monotonicity
$\dfrac{1}{n+1}\lt\dfrac1n$ for every $n$, so $b_n$ is decreasing. ✓
3
Check the limit
$\displaystyle\lim_{n\to\infty}\dfrac1n=0$. ✓
4
Conclude
The series converges by the alternating series test.
5
Note the contrast
Without the alternating signs this is the harmonic series, which diverges. The signs alone changed the verdict.
EX 2
A failure of the limit condition
the wrong testDetermine whether $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\dfrac{n}{n+1}$ converges.
1
Strip the sign
$b_n=\dfrac{n}{n+1}$.
2
Check the limit
$\displaystyle\lim_{n\to\infty}\dfrac{n}{n+1}=1\neq0$. Condition 2 fails.
3
Draw the right conclusion
A failed alternating series test proves nothing on its own.
4
Switch tests
Since $b_n\to1$, the terms $a_n=(-1)^{n}b_n$ oscillate near $\pm1$ and have no limit.
5
Conclude
By the $n$th-term test, the series diverges. Naming the correct test is part of the answer.
EX 3
Proving ‘decreasing’ with a derivative
when it is not obviousShow that $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\dfrac{\ln n}{n}$ converges.
1
Strip the sign
$b_n=\dfrac{\ln n}{n}$, positive for $n\ge2$.
2
Check the limit
$\displaystyle\lim_{n\to\infty}\dfrac{\ln n}{n}=0$ by L'Hôpital's rule. ✓
3
Test monotonicity via a function
$f(x)=\dfrac{\ln x}{x}$ gives $f'(x)=\dfrac{1-\ln x}{x^{2}}$.
4
Find where it decreases
$f'\lt 0$ when $\ln x\gt1$, that is $x\gt e$. So $b_n$ decreases for $n\ge3$ — eventually, which is enough.
5
Conclude
Both conditions hold from $n=3$ onward, so the series converges.
EX 4
Convergent but not absolutely
a preview of the distinctionFor $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{\sqrt{n}}$, determine convergence and then examine $\sum\lvert a_n\rvert$.
1
Apply the alternating series test
$b_n=\dfrac{1}{\sqrt{n}}$ decreases to $0$, so the series converges.
2
Strip the signs entirely
$\displaystyle\sum\left\lvert\dfrac{(-1)^{n}}{\sqrt{n}}\right\rvert=\sum\dfrac{1}{\sqrt{n}}$.
3
Identify that family
A $p$-series with $p=\dfrac12\le1$, which diverges.
4
Name the situation
Convergent as written, divergent in absolute value — this is conditional convergence.
5
Note what the test did not say
The alternating series test never distinguishes the two kinds. That question requires testing $\sum\lvert a_n\rvert$ separately.
What it cannot tell you
It cannot prove divergence
A failed condition means the test does not apply, not that the series diverges. If $b_n\not\to0$, cite the $n$th-term test for the divergence.
Both conditions are required
Terms going to zero is not enough; they must also decrease. State both checks explicitly on free response.
‘Eventually’ decreasing is enough
The conditions need only hold past some index. Finitely many terms never affect convergence.
It says nothing about absolute convergence
$\sum\dfrac{(-1)^{n}}{\sqrt{n}}$ converges while $\sum\dfrac{1}{\sqrt{n}}$ diverges. Testing $\sum\lvert a_n\rvert$ is a separate step.