How to Choose a Convergence Test: an 8-Step Flowchart with an Example at Every Step

The flowchart

The 8 steps with examples

Step 1 · Do the terms go to zero?

$\sum\dfrac{n}{n+1}$: the terms → 1 ≠ 0, so the series diverges at once. This test works in one direction only; if the terms → 0, go to the next step.

Step 2 · Is it geometric or a p-series?

$\sum\dfrac{5}{4^n}$ is geometric with $r = \tfrac14$ and converges · $\sum\dfrac{1}{\sqrt n}$ is a p-series with $p = \tfrac12 \le 1$ and diverges — remember the result, no computation needed.

Step 3 · Can you split it into partial fractions?

$\sum\dfrac{1}{n(n+1)} = \sum\left(\dfrac1n - \dfrac{1}{n+1}\right) = 1$. Telescoping gives both the verdict and the value of the sum.

Step 4 · Is it a ratio of polynomials or roots?

$\sum\dfrac{3n+5}{n^3-2}$: the dominant part is $\tfrac{3n}{n^3} = \tfrac{3}{n^2}$; the limit comparison with $\tfrac{1}{n^2}$ gives 3 (positive and finite), so it converges · Shortcut: it converges ⇔ (degree of denominator − degree of numerator) > 1.

Step 5 · Does it contain factorials or exponentials?

$\sum\dfrac{n!}{n^n}$: $\dfrac{a_{n+1}}{a_n} = \left(\dfrac{n}{n+1}\right)^n \to \dfrac1e < 1$, so it converges · If the limit is exactly 1, the ratio test is inconclusive.

Step 6 · Is the whole term raised to the n-th power?

$\sum\left(\dfrac{n}{2n+1}\right)^n$: $\sqrt[n]{a_n} = \dfrac{n}{2n+1} \to \tfrac12$, so it converges; the root removes the exponent immediately.

Step 7 · Does it contain ln, or is it easy to integrate?

$\sum\dfrac{1}{n\ln n}$: $\int_2^\infty\dfrac{dx}{x\ln x} = [\ln\ln x]_2^\infty = \infty$, so it diverges (even though computer sums look as if they settle) · $\sum\dfrac{1}{n(\ln n)^2}$, on the other hand, converges.

Step 8 · Do the signs alternate?

$\sum\dfrac{(-1)^n}{\sqrt n}$: the absolute values $\sum\tfrac{1}{\sqrt n}$ diverge, but $\tfrac{1}{\sqrt n}$ decreases to 0, so Leibniz says the series converges — it converges conditionally.

Summary table for exam revision

Three traps that cost marks

• $\sum\dfrac{1}{n^{1 + 1/n}}$ looks like a p-series with p > 1, but the exponent tends to 1; the limit comparison with $\tfrac1n$ gives 1, so it diverges.
• $\sum\left(1 + \tfrac1n\right)^n$ does not need the root test: the terms → e ≠ 0, so it diverges by the n-th term test.
• $\sum\dfrac{1}{3^n - n}$: direct comparison with $\tfrac{1}{3^n}$ fails because the terms are larger; use the limit comparison test instead.

Build the reflex with the 40-question test-choice game (in Thai), which checks both the test you choose and your verdict.