Choosing a Convergence Test: AP Calc BC Series Guide

Use this ordered checklist and stop at the first test that fits. First, run the nth-term test: if the terms do not shrink to 0, the series diverges and you are done. Next, recognize the form on sight: a constant ratio to the n-th power is geometric, 1 over n to the p is a p-series, and an alternating sign is the alternating series test. If you see a factorial or something raised to the n-th power, use the ratio test (or the root test for an n-th power); if it looks like a rational function of n, use limit comparison to a p-series; and if it is positive, decreasing, and easy to integrate, use the integral test. The single most useful tie-break: a factorial or an n-th power almost always means the ratio test.

You are not supposed to try tests at random. The reason series convergence feels like panic is that students pick a test by vibes, go down one road, find out it is useless, and start all over. Instead, run this ordered checklist on the series and stop at the first test that fits.

Step 1 — nth-term test, always first. Look at the terms a_n. If they do NOT shrink to 0, the series diverges immediately and you are done. (If they do go to 0, this test tells you nothing yet — keep going.)

Step 2 — recognize the FORM by sight. A constant ratio r to the n-th power is geometric (converges exactly when the absolute value of r is less than 1). A term like 1 over n to the p is a p-series (converges exactly when p is greater than 1). A (−1) to the n making terms flip sign is the alternating series test (converges if the terms decrease to 0).

Step 3 — see a factorial or an n-th power? Use the ratio (or root) test. This is the single most useful tie-break. Any n! or something raised to the n-th power screams ratio test, because the factorial or exponential cancels cleanly in the ratio a_(n+1) over a_n. For a pure something-to-the-n with no factorial, the root test is often even faster.

Step 4 — looks like a rational function of n? Use limit comparison to a p-series. If a_n behaves like a polynomial over a polynomial, compare it to 1 over n to the p, where p is the degree of the bottom minus the degree of the top. The limit-comparison ratio converging to a finite positive number means both series do the same thing.

Step 5 — positive, decreasing, and easy to integrate? Use the integral test. If a_n = f(n) for a positive decreasing f you can actually integrate (think 1 over n times ln n), the series and the improper integral converge or diverge together.

The one-line tie-breaks to memorize:

• Factorial or ^n present → ratio/root test.

• Rational in n → limit comparison to a p-series.

• Alternating sign → alternating series test.

• Recognizable r^n or 1/n^p → geometric or p-series, no work needed.

• Nothing fits and it is positive and integrable → integral test.

\text{If }a_n\not\to 0\;\Rightarrow\;\sum a_n\text{ diverges (nth-term test, run this first)}

Which convergence test should I use, and in what order?

\text{Ratio Test:}\quad L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\;\Rightarrow\; L<1\text{ converges},\; L>1\text{ diverges},\; L=1\text{ inconclusive}

Worked example: choosing a test for a factorial series

Determine whether the series, the sum from n = 1 to infinity of 2 to the n over n factorial, converges or diverges.

Step 1 — nth-term test. The terms 2^n / n! go to 0 because factorial growth beats exponential growth, so the nth-term test does not kill it. It is inconclusive, so keep going.

Step 2 — spot the form. There is a factorial in the denominator. By the checklist, a factorial means reach for the RATIO test.

Step 3 — set up the ratio of consecutive terms.

a_n = \dfrac{2^n}{n!} \qquad a_{n+1} = \dfrac{2^{n+1}}{(n+1)!}

\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n}

Step 4 — cancel. The 2^(n+1) over 2^n leaves a 2, and the n! over (n+1)! leaves 1 over (n+1).

\frac{a_{n+1}}{a_n} = \frac{2}{\,n+1\,}

Step 5 — take the limit.

L=\lim_{n\to\infty}\frac{2}{n+1}=0

Step 6 — verdict. Because L = 0 is less than 1, the ratio test says the series CONVERGES. (For a sanity check, this series in fact sums to e squared minus 1.)

Another worked example

Decide about the sum from n = 1 to infinity of n over (2n cubed plus 1). Step 1: the nth-term test is inconclusive because the terms go to 0. Step 2: this is a rational function of n, so use limit comparison to a p-series. Step 3: for large n the term behaves like n over 2n cubed, which is 1 over (2 n squared), so compare to 1 over n squared, a p-series with p = 2. Step 4: because p = 2 is greater than 1 that p-series converges, and the limit-comparison ratio is a finite positive number, so the original series converges too.

Aligned with the College Board AP Calculus BC CED — Unit 10 (Infinite Sequences and Series), Topics 10.3 through 10.9, covering the nth-term, geometric, integral, comparison, alternating, and ratio tests. College Board ↗

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