What mathematical functions can I use in the expression for aₙ?

You can use standard arithmetic operators (+, -, *, /, ^), parentheses, and functions like exp, sin, cos, sqrt, and ln with variable n.

How does the tool handle alternating series?

It detects alternating sign patterns such as cos(pi*n) or (-1)^n, checks if magnitudes decrease to zero, and applies the Leibniz alternating series test to identify conditional convergence.

Why do the ratio and root tests show 'inconclusive' for some series?

When the ratio limit L or root limit ρ equals 1 (common with p-series and rational functions), those specific tests cannot determine convergence, prompting the tool to evaluate p-series and integral tests.

What is the starting index for the series?

The evaluation assumes the summation index n starts at n = 1.

How is the total sum estimated for convergent series?

The sum is calculated by combining large-index partial sums with a tail estimate based on geometric or power-law decay.

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Series Convergence Tester (Ratio/Root/Integral/Comparison)

Numeric convergence tests for Σ aₙ: nth-term, ratio, root, p-series comparison, and integral test; alternating series classified by Leibniz with a sum estimate.

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Σ 1/n² converges — p-series with p = 2, sum ≈ 1.6449

Ratio and root tests are inconclusive for p-series (L ≈ 1, ρ ≈ 1), but the comparison estimate gives p = −ln|a(10⁹)|/ln(10⁹) = 2 > 1, so the series converges; the partial sum to 10⁵ plus the p-law tail estimates the sum as π²/6 ≈ 1.6449.

Series convergence analysis — Σ aₙ with aₙ = 1/n^2 nth-term test: |a(10^9)| ≈ 1.00e-18 → 0 — passes ratio test: L = |a(10^6+1)|/|a(10^6)| ≈ 0.999998 — inconclusive (L ≈ 1) root test: ρ = |a(10^9)|^(1/10^9) ≈ 1 — inconclusive (ρ ≈ 1) p-series comparison: p ≈ 2 — converges (p ≈ 2 > 1) (Σ 1/n^p converges for p > 1) integral test: ∫₂^10^4 |a(x)|dx ≈ 0.4999 (+tail ≈ 0.0001) — decays fast — convergent estimated sum ≈ 1.6449 (partial sum + p-law tail) VERDICT: CONVERGES (p-series comparison: p ≈ 2 > 1)

Σ cos(πn)/n converges conditionally — Leibniz alternating test

cos(πn) = (−1)ⁿ, so this is the alternating harmonic series: the signs are detected as strictly alternating and |aₙ| = 1/n decreases to 0, so the Leibniz test applies. The magnitude series behaves like p = 1 (divergent), making the convergence conditional, and the partial sum approaches −ln 2 ≈ −0.6931.

Series convergence analysis — Σ aₙ with aₙ = cos(pi*n)/n Alternating series detected (signs strictly alternate); |aₙ| decreases toward 0 — Leibniz test applies nth-term test: |a(10^9)| ≈ 1.00e-9 → 0 — passes ratio test: L = |a(10^6+1)|/|a(10^6)| ≈ 0.999999 — inconclusive (L ≈ 1) root test: ρ = |a(10^9)|^(1/10^9) ≈ 1 — inconclusive (ρ ≈ 1) p-series comparison (on |aₙ|): p ≈ 1 — diverges (p ≈ 1 ≤ 1) (borderline harmonic-like; classified with the harmonic side (p = 1 diverges)) integral test: ∫₂^10^4 |a(x)|dx ≈ 5.3924 — each decade adds a comparable amount — grows without bound — divergent estimated sum ≈ -0.6931 (Leibniz remainder bound |R| ≤ 1.0e-5) VERDICT: CONVERGES CONDITIONALLY (p-series comparison: p ≈ 1 ≤ 1; Leibniz test; not absolutely)

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ContentPaste or type the main input values.1

Term aₙ expression (variable n)textRequiredThe general term aₙ as an expression of n (n starts at 1). Supports + - * / ^, parentheses, and functions like exp, sin, cos, sqrt, ln. Alternating signs (e.g. cos(pi*n)) are detected.

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Series Convergence Tester (Ratio/Root/Integral/Comparison)

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Examples

1. Analyzing a P-Series

2. Testing an Alternating Series for Conditional Convergence

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