--- title: "Convergence Tests for Series of Positive Terms" description: "Decide convergence with the integral, comparison, limit comparison, ratio and root tests, and choose the test that matches the shape of the terms." canonical: https://lightmysky.com/learn/mathematics/convergence-tests-for-series-of-positive-terms-mt_8N49Q4BlO8 source: https://lightmysky.com/learn/mathematics/convergence-tests-for-series-of-positive-terms-mt_8N49Q4BlO8.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Convergence Tests for Series of Positive Terms Decide convergence with the integral, comparison, limit comparison, ratio and root tests, and choose the test that matches the shape of the terms. Subject: Mathematics · Area: Calculus & Analysis · Ages 19 to 20 Page: https://lightmysky.com/learn/mathematics/convergence-tests-for-series-of-positive-terms-mt_8N49Q4BlO8 ## Ready when they can - Choose and apply p-series, integral, ratio and root tests from the form of the terms - Use direct or limit comparison with a known series and justify the conclusion - Explain why a ratio limit of 1 is inconclusive and choose a follow-up argument ## Lesson: Matching each series to its test You have defined a series sum using partial sums. Here, sum a_n means the infinite sum of the terms a_n; n starts at 1 unless stated otherwise. The p-series sum 1/n^p converges exactly when p > 1 and diverges when p <= 1. Combine denominator powers: n^2 sqrt(n)=n^(5/2). Changing finitely many starting terms does not change convergence. Direct comparison requires nonnegative terms. If 0 <= a_n <= b_n eventually and sum b_n converges, sum a_n converges. If a_n >= b_n >= 0 eventually and sum b_n diverges, sum a_n diverges. For example, 0 < 1/(n^2+7) < 1/n^2 proves convergence. A convergent lower bound or a divergent upper bound gives no conclusion by itself. For limit comparison, take positive a_n and b_n. If L=lim(a_n/b_n) is finite and strictly positive, both series converge or both diverge. Use dominant powers to choose a benchmark, then calculate the limit. With a_n=(2n+1)/(n^2+4) and b_n=1/n, the ratio is (2n^2+n)/(n^2+4). Divide by n^2 to find L=2. The harmonic benchmark diverges, so sum a_n diverges. For positive terms, calculate the ratio-test limit L=lim(a_(n+1)/a_n), when it exists. L < 1 proves convergence; L > 1, including infinity, proves divergence. L=1 is inconclusive. For a p-series the ratio is (n/(n+1))^p, which tends to 1. Yet p=1 diverges and p=2 converges. Both outcomes are possible when the ratio limit is 1. For a_n=3^n/n!, use (n+1)!=(n+1)n! to simplify the consecutive ratio to 3/(n+1). Its limit 0 is below 1, proving convergence. When n^n occurs, the limit (1+1/n)^n=e can help; reciprocals tend to 1/e. Cancel within terms too: n!/(n+1)!=1/(n+1). Comparing its terms with 1/n gives limit 1 and proves divergence, although its consecutive ratio also tends to 1. For a_n=n^k/c^n, with fixed real k and c > 0, the consecutive ratio is (1+1/n)^k/c. Its limit is 1/c. Thus c > 1 proves convergence by the ratio test. For example, n/5^n gives ratio (n+1)/(5n), tending to 1/5. An exponential in a formula is a reason to try the test, not a conclusion without calculation. For nonnegative terms, the root test uses R=lim(a_n^(1/n)), when it exists. R < 1 proves convergence; R > 1, including infinity, proves divergence; R=1 is inconclusive. When a_n=(b_n)^n with b_n positive, taking the nth root leaves b_n. For [(n+2)/(4n+1)]^n, the root tends to 1/4, so the series converges. The integral test needs f positive, continuous and decreasing on [N,infinity), with f(n)=a_n. The series and the improper integral from N to infinity both converge or both diverge; their values need not agree. For f(x)=1/(x+2)^2 on x >= 1, f'(x)=-2/(x+2)^3 < 0. Its antiderivative is -1/(x+2). The integral from 1 to t is 1/3-1/(t+2), tending to 1/3. Thus the series converges; its sum is not determined. **Recap.** Check each test's conditions and calculate its limit. Ratio and root limits of 1 are inconclusive. Comparison needs a justified benchmark; an integral test does not give the series sum. *The lesson and practice material were written for this dataset using standard mathematical definitions and convergence tests.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Infinite Series and the Geometric Series](https://lightmysky.com/learn/mathematics/infinite-series-and-the-geometric-series-mt_EPE-ERFFQt) ## Optional connections - [Improper Integrals and Their Convergence](https://lightmysky.com/learn/mathematics/improper-integrals-and-their-convergence-mt_gvtZPFRA6Z) ## Opens up - [Alternating Series and Absolute Convergence](https://lightmysky.com/learn/mathematics/alternating-series-and-absolute-convergence-mt_sk6F4n1USH)