Intersect Prep

AP Calculus BC series tests: which one to use and when

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Series is one of the two heaviest units on the AP Calculus BC exam, and picking a convergence test is the part students find hardest. The tests themselves are short. The skill is knowing which one the series is asking for.

This guide gives you an order to check them in, shows each test on a worked example, and explains what to write so a free-response answer earns the point.

The order to check

  1. 1Do the terms approach 0? If not, the series diverges by the nnth term test, and you're done.
  2. 2Is it geometric or a pp-series? Then you know the answer on sight.
  3. 3Does it alternate? Try the alternating series test, then check for absolute convergence if asked.
  4. 4Does it have factorials or nnth powers? Try the ratio test.
  5. 5Is it a fraction of powers of nn? Use a limit comparison or direct comparison with a pp-series.
  6. 6Is it easy to integrate? Use the integral test.

The nth term test

If lim⁡n→∞an≠0\lim_{n \to \infty} a_n \ne 0, the series diverges. It only ever proves divergence. A limit of 00 tells you nothing, as the harmonic series shows.

Terms that don't shrink

Does ∑n=1∞3n2+15n2−n\sum_{n=1}^{\infty} \frac{3n^2 + 1}{5n^2 - n} converge or diverge?

The terms approach 35\frac{3}{5}, which isn't 00, so the series diverges by the nnth term test.

Answer: Diverges

Geometric series and p-series

  • Geometric: ∑arn\sum a r^n converges when ∣r∣<1|r| < 1, and the sum from n=0n = 0 is a1−r\frac{a}{1 - r}. It's the one kind of series whose sum you'll regularly be asked for.
  • pp-series: ∑1np\sum \frac{1}{n^p} converges when p>1p > 1 and diverges when p≤1p \le 1.

Sum of a geometric series

Find ∑n=0∞5(23)n\sum_{n=0}^{\infty} 5\left(\frac{2}{3}\right)^n.

Here a=5a = 5 and r=23r = \frac{2}{3}, so the sum is 51−2/3=15\frac{5}{1 - 2/3} = 15.

Answer: 1515

The comparison tests

Keep the highest powers of nn on top and bottom to see what the series behaves like, then compare with that simpler series. Limit comparison is usually easier because you don't need an inequality. Both tests need positive terms. For direct comparison, a series smaller than a convergent one converges, and a series larger than a divergent one diverges. For limit comparison, the limit of the ratio must be positive and finite.

Limit comparison

Does ∑n=1∞4n−1n3+7\sum_{n=1}^{\infty} \frac{4n - 1}{n^3 + 7} converge or diverge?

For large nn the terms behave like 4nn3\frac{4n}{n^3}, so compare with bn=1n2b_n = \frac{1}{n^2}. Then lim⁡n→∞(4n−1)/(n3+7)1/n2=4\lim_{n \to \infty} \frac{(4n - 1)/(n^3 + 7)}{1/n^2} = 4, which is positive and finite.

Since ∑1n2\sum \frac{1}{n^2} is a convergent pp-series, the given series converges by the limit comparison test.

Answer: Converges

Direct comparison

Does ∑n=1∞12n+n\sum_{n=1}^{\infty} \frac{1}{2^n + n} converge or diverge?

For every n≥1n \ge 1, 0<12n+n<12n0 < \frac{1}{2^n + n} < \frac{1}{2^n}. The series ∑12n\sum \frac{1}{2^n} is geometric with r=12r = \frac{1}{2}, so it converges, and therefore the smaller series converges too.

Answer: Converges

The alternating series test

An alternating series converges if the size of its terms decreases and approaches 00. Both conditions need to be stated on a free-response question.

Conditional convergence

Is ∑n=1∞(−1)n+1n2/3\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2/3}} absolutely convergent, conditionally convergent, or divergent?

The terms 1n2/3\frac{1}{n^{2/3}} are positive, decrease, and approach 00, so the series converges by the alternating series test.

510150.51
The partial sums bounce around the limit, with shrinking swings.

Without the signs, it's ∑1n2/3\sum \frac{1}{n^{2/3}}, a pp-series with p=23≤1p = \frac{2}{3} \le 1, which diverges. So the convergence is conditional.

Answer: Conditionally convergent

The ratio test

Calculate L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|. The series converges absolutely if L<1L < 1 and diverges if L>1L > 1. If L=1L = 1, the test tells you nothing, so try another one. It's the main tool for power series, too.

Factorials and powers

Does ∑n=1∞n⋅4nn!\sum_{n=1}^{\infty} \frac{n \cdot 4^n}{n!} converge or diverge?

an+1an=(n+1)4n+1(n+1)!⋅n!n⋅4n=4n\frac{a_{n+1}}{a_n} = \frac{(n+1)4^{n+1}}{(n+1)!} \cdot \frac{n!}{n \cdot 4^n} = \frac{4}{n}, which approaches L=0<1L = 0 < 1. The series converges by the ratio test.

Answer: Converges

The integral test

If ff is positive, continuous, and decreasing, then ∑f(n)\sum f(n) and ∫1∞f(x) dx\int_1^{\infty} f(x)\,dx either both converge or both diverge. Use it when the terms have an obvious antiderivative.

An easy substitution

Does ∑n=1∞n(n2+3)2\sum_{n=1}^{\infty} \frac{n}{(n^2 + 3)^2} converge or diverge?

f(x)=x(x2+3)2f(x) = \frac{x}{(x^2 + 3)^2} is positive and continuous, and its derivative is negative for x>1x > 1, so it's decreasing for x≥1x \ge 1. With u=x2+3u = x^2 + 3, ∫1∞x(x2+3)2 dx=18\int_1^{\infty} \frac{x}{(x^2 + 3)^2}\,dx = \frac{1}{8}, which is finite. The series converges by the integral test.

Answer: Converges

Using Desmos with series

Type "sum" to build a sum, and use a slider for the upper limit to watch partial sums. It's a helpful picture while you're learning, but it can't prove anything. The harmonic series grows so slowly that it can look like it's settling down, and it isn't.

Keep going

Practice on exam-style questions

Intersect Prep has 70 multiple-choice questions for Unit 10, with a full solution for each one. Create a free account to try a full lesson with the calculator. One payment of $49 unlocks the course for 6 months.

Questions

Which series tests are on the AP Calculus BC exam?
The nth term, integral, comparison, limit comparison, alternating series, and ratio tests, along with geometric series and p-series. The root test isn't part of the course.
What's the difference between absolute and conditional convergence?
A series converges absolutely if the series of absolute values converges. It converges conditionally if it converges but the series of absolute values diverges.
What happens when the ratio test gives 1?
The test is inconclusive, so use another one. This often happens with fractions of powers of n, which suit a comparison test better.
Does a limit of zero mean a series converges?
No. A limit of zero is necessary but not enough. The harmonic series has terms that approach zero, and it diverges.

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