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This cheat sheet helps students choose the right convergence test for infinite series quickly and confidently. It organizes the main tests into a decision flowchart, then connects each decision to a formula and a typical example. College calculus students need this reference because many series look similar, but small structural differences determine which test works best.

The goal is to make test selection systematic instead of based on guessing.

The most important first step is checking the term test by evaluating limnan\lim_{n \to \infty} a_n. From there, recognizable forms such as geometric series arn\sum ar^n, p-series 1np\sum \frac{1}{n^p}, alternating series, factorials, exponentials, and positive rational expressions suggest specific tests. Comparison, limit comparison, ratio, root, integral, and alternating series tests each answer different kinds of convergence questions.

A polished flowchart should guide students from the form of ana_n to the most efficient test and then to a clear conclusion.

Key Facts

  • The divergence test says if limnan0\lim_{n \to \infty} a_n \ne 0 or the limit does not exist, then an\sum a_n diverges.
  • A geometric series n=0arn\sum_{n=0}^{\infty} ar^n converges when r<1|r| < 1 and has sum a1r\frac{a}{1-r}.
  • A p-series n=11np\sum_{n=1}^{\infty} \frac{1}{n^p} converges when p>1p > 1 and diverges when p1p \le 1.
  • The integral test applies when an=f(n)a_n = f(n) and f(x)f(x) is positive, continuous, and decreasing, and an\sum a_n converges exactly when 1f(x)dx\int_1^{\infty} f(x)\,dx converges.
  • The direct comparison test says if 0anbn0 \le a_n \le b_n and bn\sum b_n converges, then an\sum a_n converges.
  • The limit comparison test says if an>0a_n > 0, bn>0b_n > 0, and limnanbn=c\lim_{n \to \infty} \frac{a_n}{b_n} = c with 0<c<0 < c < \infty, then an\sum a_n and bn\sum b_n have the same behavior.
  • The ratio test uses L=limnan+1anL = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|: the series converges if L<1L < 1, diverges if L>1L > 1, and is inconclusive if L=1L = 1.
  • The alternating series test says (1)nbn\sum (-1)^n b_n converges if bn0b_n \ge 0, bn+1bnb_{n+1} \le b_n, and limnbn=0\lim_{n \to \infty} b_n = 0.

Vocabulary

Infinite series
An infinite series is a sum of infinitely many terms, written as n=1an\sum_{n=1}^{\infty} a_n.
Convergence
A series converges if its sequence of partial sums approaches a finite number.
Divergence
A series diverges if its partial sums do not approach a finite number.
Absolute convergence
A series an\sum a_n converges absolutely if an\sum |a_n| converges.
Conditional convergence
A series an\sum a_n converges conditionally if an\sum a_n converges but an\sum |a_n| diverges.
Partial sum
The NNth partial sum is SN=n=1NanS_N = \sum_{n=1}^{N} a_n, the sum of the first NN terms.

Common Mistakes to Avoid

  • Using the divergence test to prove convergence is wrong because limnan=0\lim_{n \to \infty} a_n = 0 is necessary but not sufficient for convergence.
  • Applying the ratio test when L=1L = 1 and claiming convergence or divergence is wrong because the ratio test is inconclusive at L=1L = 1.
  • Forgetting to check positivity in comparison tests is wrong because direct and limit comparison require eventually positive terms.
  • Using the alternating series test without checking that bnb_n decreases is wrong because both bn+1bnb_{n+1} \le b_n and limnbn=0\lim_{n \to \infty} b_n = 0 are required.
  • Choosing direct comparison with the inequality in the wrong direction is wrong because anbna_n \le b_n only proves convergence from a larger convergent series, while anbna_n \ge b_n only proves divergence from a smaller divergent series.

Practice Questions

  1. 1 Determine whether n=13n5n\sum_{n=1}^{\infty} \frac{3^n}{5^n} converges, and find its sum if it converges.
  2. 2 Use an appropriate test to determine whether n=21n(lnn)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} converges.
  3. 3 Determine whether n=1n!4n\sum_{n=1}^{\infty} \frac{n!}{4^n} converges or diverges.
  4. 4 For the series n=1(1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}, explain why it converges conditionally rather than absolutely.