About the Series Convergence Calculator
This series convergence calculator tests whether an infinite series Σ a(n) converges or diverges. Type the general term as a formula in n — for example 1/n^2, (-1)^(n+1)/n, 2^n/n! or 1/(n*ln(n)) — choose the starting index, and it estimates the ratio-test limit, the root-test value and the effective power p of the terms, then applies the test that settles the question: the ratio test, a p-series or integral-test comparison, the alternating series test, or the divergence (nth-term) test.
Alongside the verdict it computes partial sums up to 10,000 terms and plots them, so you can see a convergent series level off or a divergent one keep climbing. It also reports whether convergence is absolute or only conditional, the distinction that matters for rearranging terms and for power-series endpoints.
It is designed for calculus II students checking homework and building intuition. The tests are applied to numerical estimates of the limits rather than proved symbolically, which works reliably for the standard textbook forms (powers, logarithms, exponentials, factorials and alternating signs) but irregular terms such as sin(n)/n are reported as inconclusive, and exponents extremely close to 1 (such as 1/n^1.0001) cannot be told apart from 1/n numerically; treat the result as a strong check, not a formal proof.
How to use the series convergence calculator
- 1Type the general term a(n) using n, for example 1/n^2 or (-1)^(n+1)/n.
- 2Set the starting index (use 2 for terms like 1/(n·ln n) that are undefined at n = 1).
- 3Choose how many terms to add for the partial sum.
- 4Read the verdict, the test that decided it and the reason.
- 5Check the partial-sum graph: convergent series level off, divergent ones keep moving.
Formula and method
The ratio test says Σ aₙ converges absolutely if L < 1 and diverges if L > 1; the calculator estimates L from |aₙ₊₁/aₙ| at n = 8, 16, 32, … and extrapolates (Richardson) to remove the 1/n error. When L ≈ 1 the ratio test is inconclusive, so it measures how fast the terms shrink: the local exponent p = −d ln|aₙ| / d ln n at n ≈ 10⁴ and 10⁸, fitted as p + q/ln n to detect logarithmic factors. Limit comparison with Σ 1/nᵖ then decides: p > 1 converges, p < 1 diverges, and at p = 1 the integral test on 1/(n·lnᵠ n) converges only for q > 1.
If the terms alternate in sign and decrease to 0 while Σ|aₙ| diverges, the alternating series (Leibniz) test gives conditional convergence. If the terms do not tend to 0 at all, the series diverges by the nth-term test. Partial sums are added with compensated (Kahan) summation in double precision.
- aₙ
- General term of the series
- L
- Limit of the ratio |aₙ₊₁/aₙ|
- p
- Effective power: |aₙ| ≈ C/nᵖ for large n
- Sₙ
- Partial sum of the first N terms
Worked examples
p-series Σ 1/n²
The ratio |aₙ₊₁/aₙ| tends to 1, so the ratio test is inconclusive. The terms shrink like 1/n² with p = 2 > 1, so the series converges. The first 10,000 terms sum to 1.644834, approaching π²/6 ≈ 1.644934.
Harmonic series Σ 1/n
Even though 1/n → 0, the terms behave like 1/nᵖ with p = 1, and the harmonic series diverges. Its partial sums grow like ln N + 0.5772: about 9.79 after 10,000 terms, without bound.
Alternating harmonic series Σ (−1)ⁿ⁺¹/n
The signs alternate and 1/n decreases to 0, so the alternating series test gives convergence — to ln 2 ≈ 0.693147. The series of absolute values is the harmonic series, which diverges, so convergence is only conditional.
Factorial in the denominator: Σ 2ⁿ/n!
aₙ₊₁/aₙ = 2/(n + 1) → 0 < 1, so the ratio test shows absolute convergence. The sum from n = 1 is e² − 1 ≈ 6.389056.
Borderline: Σ 1/(n·ln n) from n = 2
These terms shrink slightly faster than 1/n, but ∫ dx/(x ln x) = ln(ln x) grows without bound, so the integral test shows divergence. The partial sum after 10,000 terms is only about 3.02 — divergence this slow is invisible in the numbers alone.
Terms that do not go to zero: Σ n/(n+1)
n/(n + 1) → 1, not 0, so the nth-term test shows divergence immediately. Each term adds almost 1, and 1,000 terms already sum to about 993.5.
Frequently asked questions
How do you know if a series converges or diverges?+
First check that the terms go to 0 (if not, it diverges). Then use the ratio or root test for exponentials and factorials, a p-series or limit comparison for powers of n, the integral test for logarithmic terms, and the alternating series test when signs alternate.
What does the ratio test say?+
Compute L = lim |aₙ₊₁/aₙ|. If L < 1 the series converges absolutely, if L > 1 (or infinite) it diverges, and if L = 1 the test is inconclusive — as for every p-series, which is why other tests are needed.
What is the difference between absolute and conditional convergence?+
A series converges absolutely if Σ|aₙ| converges. It converges conditionally if Σ aₙ converges but Σ|aₙ| does not, like the alternating harmonic series. Conditionally convergent series can be rearranged to sum to any value.
When does a p-series converge?+
The p-series Σ 1/nᵖ converges when p > 1 and diverges when p ≤ 1. So Σ 1/n² converges (to π²/6), while Σ 1/n and Σ 1/√n diverge.
If the terms go to zero, does the series converge?+
Not necessarily. Terms tending to 0 is required but not sufficient: the harmonic series Σ 1/n has terms that go to 0 yet its partial sums grow without bound, roughly like ln n.
Can a computer prove that a series converges?+
Numerical partial sums alone cannot prove convergence — Σ 1/(n·ln n) diverges but is only about 3 after 10,000 terms. This tool therefore estimates the limits the standard tests use, which is far more reliable, but it is still a numerical check rather than a proof.