Section 2.1 Summary of Series
The culmination of this chapter is to be able to take a given series, and determine if it converges or diverges. For each of these kinds of questions, you should:
- State convergence or divergence, and also,
- Specify which test you used, and give a justification for why it can be used (which is how you show your work).
For most questions, the test you should use will not be given, so you will have to develop intuition to recognize which test to use.
Subsection 2.1.1 Choosing Which Test to Use
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Do the terms not approach 0?Use the \(n\)th term test for divergence: If \(\lim_{n \to \infty} a_n \neq 0\) (or the limit doesn’t exist), then the series diverges immediately.
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Are there exponents of \(n\text{?}\)Could be a geometric series, of the form \(\sum a\,r^n\text{.}\)Use the geometric series test: For common ratio \(r\text{,}\)
- \(|r| \lt 1\) ⇒ converges
- \(|r| \geq 1\) ⇒ diverges
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Are two similar terms being subtracted? Or, can you do partial‑fraction decomposition?Could be telescoping.Write the partial sum \(s_n\) explicitly, then take \(\lim_{n \to \infty} s_n\text{.}\)
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Can it be written as a power of \(n\text{?}\)Consider the p‑series \(\sum \frac{1}{n^{p}}\)\(p\)‑series: converges if \(p \gt 1\text{,}\) diverges if \(p \leq 1\text{.}\)
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Does the numerator or denominator have more than one term?Consider asymptotic comparison (keeping only the dominant term)Use the direct or limit comparison test: compare to a simpler series (typically a p‑series or geometric series).
- For limit comparison, calculate \(L = \lim_{n \to \infty} \frac{a_n}{b_n}\text{.}\) If \(L \gt 0\text{,}\) then \(\sum a_n\) and \(\sum b_n\) behave the same.
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Has factorials (\(n!\)) and/or exponential expressions (\(b^n\))?Use the ratio test: calculate \(\rho = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}\)
- \(\rho \lt 1\) ⇒ converges (absolutely)
- \(\rho \gt 1\) ⇒ diverges
- \(\rho = 1\) ⇒ inconclusive (use another test)
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Has a double power (like \(n^n\) or \((f(n))^n\))?Use the root test: calculate \(L = \lim_{n \to \infty} \sqrt[n]{\lvert a_n\rvert}\)
- \(L \lt 1\) ⇒ converges
- \(L \gt 1\) ⇒ diverges
- \(L = 1\) ⇒ inconclusive (use another test)
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Has an alternating factor, like \((-1)^n\) or \((-1)^{n+1}\text{?}\)Use the alternating series test: if \(\lim_{n \to \infty} a_n = 0\) and \(a_n\) is decreasing, then the series converges (conditionally).
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Terms can be integrated easily, maybe using \(u\)-substitution?Use the integral test: convergence of \(\sum a_n\) is the same as \(\int f(x) \,dx\)
