Introduction to Sequences & Series
In AP Calculus BC, sequences are ordered lists of numbers: $a_1, a_2, a_3, \dots$ and a series is the sum of the terms of a sequence: $ \sum_{n=1}^{\infty} a_n. $
The partial sum is $ S_n = a_1 + a_2 + \dots + a_n. $ We say the series converges if $ \lim_{n\to\infty} S_n $ exists and is finite.
Below is a simple visualization of the partial sums for a Geometric Series with $ r = 0.5. $
The line chart shows how the partial sums approach the infinite geometric series sum.
n-th Term Test (AP Calculus BC)
Statement: If $ \lim_{n \to \infty} a_n \neq 0, $ then the series $ \sum a_n $ diverges. If $ \lim_{n \to \infty} a_n = 0, $ the test is inconclusive and we must use another test.
Consider $ \sum_{n=1}^{\infty} \frac{n}{n+1}. $ Each term $ a_n = \frac{n}{n+1} \to 1 \neq 0 $ as $ n \to \infty. $ By the n-th Term Test, this series diverges.
Consider $ \sum_{n=1}^{\infty} \frac{1}{n}. $ Each term $ a_n = \frac{1}{n} \to 0 $ as $ n \to \infty. $ The test is inconclusive; we rely on the fact that $ \sum \frac{1}{n} $ (the harmonic series) diverges.
Direct Comparison Test
Statement: If $ 0 \le a_n \le b_n $ for sufficiently large $ n $ and $ \sum b_n $ converges, then $ \sum a_n $ converges. If $ a_n \ge b_n \ge 0 $ and $ \sum b_n $ diverges, then $ \sum a_n $ diverges.
Consider $ \sum_{n=1}^{\infty} \frac{1}{n^2 + 1}. $ For $ n \ge 1, $ $ \frac{1}{n^2 + 1} \le \frac{1}{n^2}. $ Since $ \sum \frac{1}{n^2} $ converges (p-series with $ p = 2 $), by Direct Comparison, our given series also converges.
Consider $ \sum_{n=4}^{\infty} \frac{1}{\sqrt{n} - 1}. $ For $ n \ge 4, $ $ \frac{1}{\sqrt{n} - 1} \ge \frac{1}{\sqrt{n}}. $ Since $ \sum \frac{1}{\sqrt{n}} $ diverges (p-series with $ p = \frac{1}{2} < 1 $), by Direct Comparison, our given series also diverges.
Limit Comparison Test
Statement: If $ a_n > 0 $ and $ b_n > 0 $, and $ \lim_{n \to \infty} \frac{a_n}{b_n} = L, $ where $ 0 < L < \infty, $ then $ \sum a_n $ and $ \sum b_n $ either both converge or both diverge.
Consider $ \sum_{n=1}^{\infty} \frac{2n + 1}{n^2}. $ Compare with the harmonic series $ \sum \frac{1}{n}. $ Then $ \lim_{n \to \infty} \frac{\frac{2n+1}{n^2}}{\frac{1}{n}} = \lim_{n \to \infty} \left(\frac{2n+1}{n^2} \cdot n\right) = 2. $ Since $ \sum \frac{1}{n} $ diverges, $ \sum \frac{2n+1}{n^2} $ diverges by Limit Comparison.
Consider $ \sum_{n=1}^{\infty} \frac{1}{n^2 - n}. $ Compare with $ \sum \frac{1}{n^2}. $ $ \lim_{n \to \infty} \frac{\frac{1}{n^2 - n}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n^2 - n} = 1. $ Since $ \sum \frac{1}{n^2} $ converges, $ \sum \frac{1}{n^2 - n} $ converges by Limit Comparison.
p-Series Test
p-series: $ \sum_{n=1}^{\infty} \frac{1}{n^p}. $ This series converges if $ p > 1 $ and diverges if $ p \le 1. $
1. $ \sum \frac{1}{n^2} \quad (p=2 > 1) \implies \text{convergent}. $
2. $ \sum \frac{1}{n} \quad (p=1) \implies \text{divergent (harmonic)}. $
3. $ \sum \frac{1}{n^{1.1}} \quad (p=1.1 > 1) \implies \text{convergent}. $
4. $ \sum \frac{1}{\sqrt{n}} \quad (p=\tfrac{1}{2} < 1) \implies \text{divergent}. $
Alternating Series Test
Statement: For $ \sum_{n=1}^{\infty} (-1)^n b_n, $ if $ b_n > 0, $ is decreasing, and $ \lim_{n \to \infty} b_n = 0, $ then the series converges.
The alternating harmonic series $ \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} $ satisfies these conditions, so it converges.
Consider $ \sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1}. $ Here, $ \lim_{n \to \infty} \frac{n}{n+1} = 1 \neq 0, $ so the series diverges (the terms do not go to zero).
Ratio Test
Statement: For $ \sum a_n, $ define
$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. $
- If $ L < 1, $ the series converges absolutely.
- If $ L > 1, $ the series diverges.
- If $ L = 1, $ the test is inconclusive.
Consider $ \sum_{n=1}^{\infty} \frac{n}{2^n}. $ Let $ a_n = \frac{n}{2^n}. $ Then $ \left|\frac{a_{n+1}}{a_n}\right| = \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} = \frac{n+1}{2n} \to \frac{1}{2} < 1. $ Hence, it converges by the Ratio Test.
Root Test
Statement: For $ \sum a_n, $ define
$ L = \lim_{n \to \infty} \sqrt[n]{|a_n|}. $
- If $ L < 1, $ the series converges absolutely.
- If $ L > 1, $ the series diverges.
- If $ L = 1, $ the test is inconclusive.
Consider $ \sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^n. $ Then $ \sqrt[n]{\left(\frac{n}{2n+1}\right)^n} = \frac{n}{2n+1} \to \frac{1}{2} < 1. $ Hence, it converges by the Root Test.
Integral Test
Statement: If $ a_n = f(n) $ for a function $ f $ that is positive, continuous, and decreasing (for $ n \ge M $), then $ \sum_{n=M}^{\infty} a_n $ converges if and only if the improper integral $ \int_{M}^{\infty} f(x)\,dx $ converges.
Consider $ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}. $ Define $ f(x) = \frac{1}{x(\ln x)^2}, $ which is positive, continuous, and decreasing for $ x \ge 2. $ Evaluate $ \int_{2}^{\infty} \frac{1}{x(\ln x)^2} \, dx. $ A known antiderivative is $ \int \frac{1}{x(\ln x)^2} \, dx = -\frac{1}{\ln x} + C. $ Evaluating from $ 2 $ to $ \infty $: $ \lim_{t \to \infty} \left[-\frac{1}{\ln x}\right]_{2}^{t} = \lim_{t \to \infty} \left(-\frac{1}{\ln t} + \frac{1}{\ln 2}\right) = 0 + \frac{1}{\ln 2}. $ Since the integral converges to a finite value, the series $ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} $ converges by the Integral Test.