Limit Comparison Test Calculator: How to Prove a Series Converges or Diverges
About the Author
Marko Šinko
Co-Founder & Lead Developer, AI Math Calculator
A limit comparison test calculator settles one question: does your series Σaₙ live or die with a simpler benchmark series Σbₙ whose fate you already know? Compute c = lim aₙ/bₙ. If that limit is a positive finite number, the two series stand or fall together — no inequalities to wrestle, no clever bounding tricks. The tool above runs the computation; this page covers the judgment calls it can’t make for you. Which bₙ to pick. What the three cases of c actually let you conclude. And — the question every Calc 2 student eventually asks — when you should reach for a completely different test instead.
Where Direct Comparison Stalls — and LCT Doesn’t
Try to prove Σ 1/(n² − n) converges by direct comparison. The natural benchmark is Σ 1/n², a convergent p-series. But 1/(n² − n) is bigger than 1/n² for every n ≥ 2 — a smaller denominator makes a larger fraction. Direct comparison needs your terms below a convergent series, and here the inequality points the wrong way. You can rescue the argument by comparing against 2/n² and proving the bound holds for n ≥ 2, but that’s exactly the kind of fiddly bookkeeping that eats exam time.
The limit comparison test skips all of it. Compute the limit of the ratio: (1/(n² − n)) ÷ (1/n²) = n²/(n² − n) = 1/(1 − 1/n) → 1. The limit c = 1 is positive and finite, Σ 1/n² converges, so Σ 1/(n² − n) converges. One limit replaces the entire inequality hunt. That’s the trade LCT offers: it only cares how aₙ and bₙ compare asymptotically, not term by term.
The Three Cases: What Each Value of c Licenses
Suppose aₙ > 0 and bₙ > 0, and let c = lim aₙ/bₙ. The limit comparison test splits into three cases, and they are not symmetric — two of them only work in one direction:
| Limit c | If Σbₙ converges | If Σbₙ diverges |
|---|---|---|
| 0 < c < ∞ | Σaₙ converges | Σaₙ diverges |
| c = 0 | Σaₙ converges | No conclusion |
| c = ∞ | No conclusion | Σaₙ diverges |
The asymmetry makes sense once you read c as a size comparison. c = 0 says aₙ is eventually far smaller than bₙ — being smaller than a convergent series forces convergence, but being smaller than a divergent one tells you nothing. Flip the logic for c = ∞. Only the middle case, 0 < c < ∞, transfers information in both directions, which is why a well-chosen bₙ — one that gives a finite nonzero c — is worth thirty seconds of thought.
Picking bₙ: Keep the Dominant Terms, Drop the Rest
The recipe: in the numerator and denominator, keep only the fastest-growing term, then simplify. For a rational term like aₙ = (3n² + 5)/(n⁴ + 2n), the bullies are 3n² on top and n⁴ below, so aₙ behaves like 3n²/n⁴ = 3/n² — take bₙ = 1/n² and expect c = 3. Degree of denominator minus degree of numerator gives the exponent: here 4 − 2 = 2, and since 2 > 1 the series converges. Roots count as fractional powers: 1/√(n³ + 4n) behaves like 1/n^(3/2), a convergent p-series. An exponential term dominates every power of n, so for 1/(2ⁿ − n) the right benchmark is the geometric series (1/2)ⁿ, not any p-series. If you need a refresher on which benchmarks converge in the first place, the series convergence calculator covers p-series and geometric series behavior in detail.
This dominant-term habit is exactly what the “Suggest bₙ for me” button in the calculator automates: it measures how fast aₙ decays numerically — a log-log slope for power-law decay, a term-to-term ratio for geometric decay — and proposes the matching benchmark.
LCT, Direct Comparison, Ratio, or Integral? A 20-Second Decision
Here’s a fact that surprises students: the ratio test returns L = 1 — inconclusive — for every rational function of n. Try it on 1/n²: the ratio of consecutive terms is n²/(n+1)² → 1. All that work for nothing. Each convergence test has a home turf:
| Test | Reach for it when aₙ has… | It breaks down when… |
|---|---|---|
| Limit comparison | Polynomials, roots, anything that behaves like nᵖ | No clean benchmark exists (e.g. 1/(n ln n)) |
| Direct comparison | An obvious inequality in the right direction | The inequality points the wrong way |
| Ratio test | Factorials or exponentials (n!, 2ⁿ) | Rational terms — always returns L = 1 |
| Root test | Whole expressions raised to the nth power | No nth-power structure to exploit |
| Integral test | Log factors like 1/(n ln n), decreasing f(x) | The antiderivative is hard or unavailable |
Rational and root-heavy terms are LCT territory. Factorials and exponentials belong to the ratio test calculator; expressions of the form (something)ⁿ suit the root test calculator; and logarithm-laced terms like 1/(n ln n) — where no p-series benchmark produces a finite nonzero c — fall to the integral test calculator. If you’re staring at a series with no idea where to start, the convergence test calculator walks through test selection automatically.
Two Complete Runs with the Limit Comparison Test Calculator
Run 1: Σ (2n + 1)/(n² + 4). Dominant terms: 2n over n², so aₙ behaves like 2/n. Take bₙ = 1/n. The ratio is (2n + 1)/(n² + 4) × n = (2n² + n)/(n² + 4), and dividing top and bottom by n² gives (2 + 1/n)/(1 + 4/n²) → 2. So c = 2, positive and finite. The benchmark Σ 1/n is the harmonic series, which diverges — slowly, but certifiably (its partial sums pass 10 only around n ≈ 12,367). By the middle row of the case table, Σ (2n + 1)/(n² + 4) diverges.
Run 2: Σ sin(1/n). No polynomial in sight, but for small x, sin x ≈ x — that’s the first term of its Taylor series. Since 1/n → 0, the terms sin(1/n) should behave like 1/n. Take bₙ = 1/n: c = lim sin(1/n)/(1/n) = 1, the classic limit sin(x)/x → 1 in disguise. Positive, finite, benchmark diverges — so Σ sin(1/n) diverges. Punch this one into the calculator above and watch the ratio column march toward 1: it’s 0.99833 already at n = 10.
The Fine Print: Positive Terms and the Cost of a Sloppy Benchmark
Two hypotheses guard the test. First, both series need positive terms — at least from some index onward. A series like Σ (−1)ⁿ/n breaks the machinery entirely; ratios of sign-flipping terms don’t have a meaningful limit. Test the absolute values instead, or hand it to the alternating series calculator. Second, you must already know what Σbₙ does. The whole test is a transfer of verdicts — pick a bₙ you can’t classify and you’ve transferred nothing.
A sloppy benchmark doesn’t give a wrong answer; it gives no answer. Compare aₙ = 1/n² against bₙ = 1/n and you get c = lim n/n² = 0 with a divergent benchmark — the dead cell of the case table. The series converges perfectly well; your comparison just failed to prove it. Undershooting is the mirror image: aₙ = 1/√n against bₙ = 1/n² gives c = ∞ against a convergent benchmark, again nothing. The fix is always the same — re-read the dominant terms and match the decay rate exactly. When the ratio column in the calculator sprints toward 0 or ∞ instead of flattening out, that’s the tool telling you your benchmark missed the growth rate, not that the test failed.
