Series Convergence Calculator
series convergence calculator. Get instant, accurate results. Enter values for instant results with step-by-step formulas.
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Formula: Ratio Test: L = |a(n+1)/a(n)| → if L<1 converges, L>1 diverges, L=1 inconclusive
Worked example — Converges. Partial sum → 1.0
Ratio Test: L = |a(n+1)/a(n)| → if L<1 converges, L>1 diverges, L=1 inconclusive
Tests for series convergence: Ratio test computes the limit of consecutive term ratios. Also checks: partial sums trend, term magnitude approaching zero (necessary condition), and root test.
Problem:a(n) = 1/2^n, starting at n=1
Solution:Ratio = 0.5 < 1, converges. Sum = 1/(1-0.5) = 1 (starting at n=0) or 1 (starting at n=1)
Result:Converges. Partial sum → 1.0
Ratio test, root test, comparison test, integral test, alternating series test, p-series test (converges if p>1), and geometric series test (converges if |r|<1).
No. If lim a(n) ≠ 0, the series MUST diverge (nth term test). However, terms approaching zero is necessary but NOT sufficient — the harmonic series 1/n diverges despite 1/n → 0.
The ratio test is inconclusive when L = 1, which often occurs with p-series and polynomial-over-polynomial terms. In these cases, use the p-series test (Σ 1/nᵖ converges iff p > 1), the comparison test against a known series, or the integral test. For alternating series like Σ (-1)ⁿ/n, use the alternating series test instead.
A series converges absolutely if Σ |aₙ| converges. It converges conditionally if Σ aₙ converges but Σ |aₙ| diverges. The alternating harmonic series Σ (-1)ⁿ/n is a classic example: it converges conditionally but not absolutely. Absolutely convergent series can be rearranged without changing the sum; conditionally convergent series cannot.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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