--- title: "How to Choose a Convergence Test: an 8-Step Flowchart with an Example at Every Step" url: https://modsengineeringmath.com/en/blog/convergence-test-flowchart/ site: "Mod's Engineering Mathematics (อาจารย์มด)" language: en updated: 2026-10-24 --- > Pick the right convergence test the first time: an 8-step flowchart (n-th term, geometric, p-series, telescoping, limit comparison, ratio, root, integral, alternating) with examples and traps. Markdown copy of https://modsengineeringmath.com/en/blog/convergence-test-flowchart/ for AI assistants and search tools. Math is LaTeX ($…$ and $$…$$); interactive labs and figures are only on the web page. # How to Choose a Convergence Test: an 8-Step Flowchart with an Example at Every Step THภาษาไทยENEnglish // Article · Sequences & series In a Calculus 2 exam, the students who finish the series questions quickly are not the ones who memorised the most theorems; they are the ones who look at a term and know at once which tool to use. This article turns that experience into an 8-step flowchart, ordered from the fastest check to the one that takes the most work, with one example per step. In 30 seconds - Always check the n-th term first (5 seconds), then ask whether it is geometric or a p-series - Ratio of polynomials → limit comparison · factorials/exponentials → ratio · whole term to the n-th power → root · contains ln → integral - Alternating signs: test Σ|aₙ| first, then use Leibniz ## The flowchart Read from top to bottom, stop at the first “yes” and use the test on the right. ## 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 | Test | Use when | Inconclusive when | | n-th term | Every question (first check) | Terms → 0 | | Limit comparison | Ratios of polynomials, roots, sin(1/n) | The limit is 0 or ∞ in the wrong direction | | Ratio | n!, cⁿ, nⁿ | ρ = 1 | | Root | The whole term to the n-th power | ρ = 1 | | Integral | ln n, functions you can integrate | f is not decreasing | | Alternating series | (−1)ⁿ bₙ | bₙ is not decreasing | ## 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. ## Continue in the library (in Thai) - Convergence tests in full: theorems, examples, Raabe and Stirling - Alternating series: Leibniz, error bound, rearrangement - Interactive series tester - Calculus 2 practice set B ## Frequently asked questions Why not use the ratio test for everything? Because series built from ratios of polynomials or roots, such as $\sum 1/n^2$ and $\sum 1/n$, always give a ratio limit of 1, which is inconclusive. The ratio test is the right tool only when the terms contain factorials or exponentials. Can the n-th term test prove convergence? No. It works in one direction only: if the terms do not tend to zero, the series diverges. If the terms do tend to zero, you must keep testing. How do I choose b(n) for the limit comparison test? Keep only the dominant (highest-power) part of the numerator and denominator. For $\dfrac{3n+5}{n^3-2}$ keep $\dfrac{3n}{n^3} = \dfrac{3}{n^2}$, so choose $b_n = 1/n^2$. Where do I start with an alternating series? Test the absolute values $\sum|a_n|$ first. If that converges, the series converges absolutely. If it diverges, apply the alternating series (Leibniz) test; if that passes, the series converges conditionally. ## More articles worth reading Sequences & series What Is a Laurent Series and How Is It Different from a Taylor Series? 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For $\dfrac{3n+5}{n^3-2}$ keep $\dfrac{3n}{n^3} = \dfrac{3}{n^2}$, so choose $b_n = 1/n^2$. **Where do I start with an alternating series?** Test the absolute values $\sum|a_n|$ first. If that converges, the series converges absolutely. If it diverges, apply the alternating series (Leibniz) test; if that passes, the series converges conditionally.