Does the Direct Comparison Test require knowing whether the comparison series converges or diverges first?

Yes. You must already know the convergence behaviour of the comparison series bₙ before you can draw conclusions about aₙ.

What value of the limit L = lim(aₙ/bₙ) makes the Limit Comparison Test inconclusive?

L = 0 or L = ∞ makes the LCT inconclusive (the test is only decisive when 0 < L < ∞).

Comparison Tests for Series — Direct & Limit

§ 01The Big Idea — Learning from a Known Neighbour

Both comparison tests share a single insight: if you already know whether a series converges or diverges, and you can sandwich an unknown series beside it, you can transfer that knowledge.

Direct convergence criteria like the Ratio Test or Root Test work beautifully for series involving factorials or exponentials. But what about a series like Σ 1/(n² + 3n + 1)? None of those power tests applies neatly. The comparison tests were designed precisely for this situation, series whose terms look like a rational function of n, or whose behaviour is close to a well-understood benchmark.

The fundamental logic is the same as squeezing. If every term of a series is smaller than the corresponding term of a series we know converges, our series must also converge, there simply is not enough "weight" in the terms to allow the sum to blow up. Conversely, if every term exceeds a divergent benchmark, the sum has no chance of being finite.

Prerequisites for this page You should be comfortable with the following before proceeding: what it means for a series to converge or diverge, the p-series test (Σ 1/nᵖ converges iff p > 1), geometric series (Σ arⁿ converges iff |r| < 1), and the nth-term divergence test. Links to all of these are in the Related Pages section.

§ 02The Comparison Arsenal — Series You Already Know

The comparison tests are only as powerful as your library of known series. These are your main benchmarks: commit them to memory.

Convergent

p-Series (p > 1)

Σ 1/nᵖ e.g. Σ 1/n²

Converges for any p > 1. Σ 1/n² = π²/6.

Divergent

Harmonic Series (p = 1)

Σ 1/n

Diverges despite terms → 0. The classic trap.

Convergent

Geometric (|r| < 1)

Σ arⁿ = a/(1−r)

e.g. Σ (1/2)ⁿ = 1. Diverges if |r| ≥ 1.

Divergent

p-Series (p ≤ 1)

Σ 1/n^p, p ≤ 1

Includes harmonic (p=1) and slower: Σ 1/√n.

Convergent

Convergent Rational

Σ 1/n(n+1)

Telescopes to 1. Also dominated by Σ 1/n².

Divergent

Geometric (r ≥ 1)

Σ rⁿ, r ≥ 1

Terms do not go to zero — diverges by nth-term test.

§ 03The Direct Comparison Test (DCT)

If you can place your series directly above or below a known one, with every single term satisfying the inequality, you are done.

Direct Comparison Test — Statement

Let 0 ≤ aₙ ≤ bₙ for all n ≥ N (for some starting index N). Then:

If Σ bₙ converges ⟹ Σ aₙ converges If Σ aₙ diverges ⟹ Σ bₙ diverges

The inequality only needs to hold from some point onwards: finitely many "bad" terms at the start do not affect convergence.

The logic is clean. If every term of your series aₙ is no bigger than the term bₙ of a convergent series, then the partial sums of Σaₙ are forever bounded above by Σbₙ: they can never exceed a finite ceiling. A bounded, non-decreasing sequence of partial sums must converge.

The reverse direction works by contrapositive. If your series aₙ is no smaller than bₙ, and Σbₙ already diverges to infinity, then Σaₙ is at least as large and must also diverge.

The two failure modes of the DCT The DCT tells you nothing in the "wrong" directions: if Σaₙ converges and aₙ ≤ bₙ, you cannot conclude anything about Σbₙ (bₙ could go either way). Likewise if Σbₙ diverges and aₙ ≥ bₙ, you know Σaₙ diverges — but this is just the same rule restated. Always set up the inequality so the known series is the one doing the "bounding" in the favourable direction.

Step-by-Step Method for the DCT

When you pick up a series and suspect DCT, here is the systematic approach:

Simplify aₙ. Drop low-order terms in numerator and denominator to guess the dominant behaviour.
Identify a candidate bₙ from your arsenal that has the same leading-term behaviour.
Prove the inequality aₙ ≤ bₙ (or aₙ ≥ bₙ) rigorously, for all n beyond some N.
State what you know about Σbₙ (converges/diverges, by which test).
Apply DCT and conclude about Σaₙ.

Worked Examples — Direct Comparison Test

Example 1 — DCT to prove convergence

Determine whether Σ_n=1^∞ 1/(n² + 3) converges or diverges.

Guess the behaviour: For large n, n² + 3 ≈ n², so aₙ ≈ 1/n². Suspect convergence.
Choose bₙ = 1/n². We know Σ 1/n² converges (p-series, p = 2 > 1).
Verify the inequality: Since n² + 3 > n² for all n ≥ 1: 1/(n² + 3) < 1/n² for all n ≥ 1 ✓
Apply DCT: 0 ≤ aₙ ≤ bₙ and Σ bₙ converges ⟹ Σ aₙ converges.
Converges

Example 2 — DCT to prove divergence

Determine whether Σ_n=1^∞ 1/(n − ln n) converges or diverges. (Assume n ≥ 2.)

Behaviour: For large n, ln n grows much slower than n, so n − ln n ≈ n. Suspect divergence (like the harmonic series).
Choose bₙ = 1/(2n) as our divergent benchmark. Σ 1/(2n) = (1/2) Σ 1/n diverges.
Verify aₙ ≥ bₙ: Need n − ln n ≤ 2n, i.e. − ln n ≤ n, i.e. ln n ≥ −n. True for all n ≥ 1 ✓. So 1/(n − ln n) ≥ 1/(2n) for all n ≥ 2 ✓
Apply DCT: 0 ≤ bₙ ≤ aₙ and Σ bₙ diverges ⟹ Σ aₙ diverges.
Diverges

Example 3 — DCT with an exponential bound

Show that Σ_n=0^∞ sin²(n)/2ⁿ converges.

Key observation: sin²(n) ≤ 1 for all n (since |sin(n)| ≤ 1).
Therefore: sin²(n)/2ⁿ ≤ 1/2ⁿ = (1/2)ⁿ for all n ≥ 0 ✓
The comparison: Σ (1/2)ⁿ is a geometric series with r = 1/2 < 1 ⟹ it converges.
Apply DCT: 0 ≤ aₙ ≤ (1/2)ⁿ and Σ (1/2)ⁿ converges ⟹ Σ sin²(n)/2ⁿ converges.
Converges

Example 4 — DCT with a careful inequality

Does Σ_n=1^∞ 1/√(n³ + 1) converge?

Behaviour: √(n³ + 1) ≈ √(n³) = n^(3/2) for large n. Expect convergence (like p = 3/2 > 1).
Choose bₙ = 1/n^(3/2). p = 3/2 > 1, so Σ 1/n^(3/2) converges.
Inequality: n³ + 1 > n³, so √(n³ + 1) > n^(3/2), giving: 1/√(n³ + 1) < 1/n^(3/2) for all n ≥ 1 ✓
Apply DCT: Σ 1/n^(3/2) converges ⟹ Σ 1/√(n³ + 1) converges.
Converges

§ 04The Limit Comparison Test (LCT)

When the inequality in the DCT is hard to establish, or you need to compare series going in the same direction, the Limit Comparison Test replaces the termwise inequality with a limit of ratios.

Limit Comparison Test — Statement

Let aₙ > 0 and bₙ > 0 for all sufficiently large n. Compute:

L = lim_n→∞ aₙ / bₙ If 0 < L < ∞ ⟹ Σaₙ and Σbₙ both converge or both diverge If L = 0 ⟹ Σbₙ converges ⟹ Σaₙ converges If L = ∞ ⟹ Σbₙ diverges ⟹ Σaₙ diverges

The decisive case is 0 < L < ∞. If aₙ/bₙ → a finite positive constant, the series share the same fate.

Why Does This Work?

If L = lim aₙ/bₙ is a finite positive number, then for large n the two sequences are proportional: aₙ ≈ L · bₙ. A convergent series scaled by a constant is still convergent; a divergent series scaled by a positive constant is still divergent. The LCT formalises this proportionality.

The edge cases L = 0 and L = ∞ are one-directional. If L = 0 then aₙ is eventually much smaller than bₙ, so convergence of Σbₙ forces convergence of Σaₙ (like a weak version of DCT). But if Σbₙ diverges and L = 0, we cannot conclude anything about Σaₙ.

LCT versus DCT — when to use which Use the DCT when the inequality aₙ ≤ bₙ (or aₙ ≥ bₙ) is easy to see directly — for example when you have a "+c" term making the denominator larger. Use the LCT when establishing a strict inequality is messy or impossible, but you can clearly identify the dominant term of aₙ. In practice, the LCT is faster for most rational-function-style series.

Step-by-Step Method for the LCT

Identify the dominant term of aₙ by keeping only the highest-power terms in numerator and denominator. Call the resulting simplified expression bₙ.
Classify Σbₙ using a known test (p-series, geometric, etc.).
Compute L = lim aₙ/bₙ as n → ∞. Simplify by dividing top and bottom by the highest power of n.
Check: is 0 < L < ∞? If yes, Σaₙ and Σbₙ have the same convergence behaviour. Apply it.
State the conclusion explicitly.

Worked Examples — Limit Comparison Test

Example 5 — LCT, rational terms

Determine whether Σ_n=1^∞ (3n² + 2n)/(n⁴ − 5) converges or diverges.

Dominant terms: Numerator ~ 3n², denominator ~ n⁴. So aₙ ~ 3n²/n⁴ = 3/n².
Choose bₙ = 1/n². Σ 1/n² converges (p = 2 > 1).
Compute L: L = lim (3n² + 2n)/(n⁴ − 5) ÷ (1/n²) = lim n²(3n² + 2n)/(n⁴ − 5) = lim (3n⁴ + 2n³)/(n⁴ − 5) Divide by n⁴: = lim (3 + 2/n)/(1 − 5/n⁴) = 3/1 = 3
L = 3, so 0 < L < ∞. LCT applies: Σaₙ and Σbₙ share the same behaviour.
Since Σ 1/n² converges ⟹ Σ (3n² + 2n)/(n⁴ − 5) converges. Converges

Example 6 — LCT proving divergence

Test Σ_n=1^∞ (√n + 1)/(n + 3) for convergence.

Dominant terms: Numerator ~ √n = n^(1/2), denominator ~ n. So aₙ ~ n^(1/2)/n = n^(−1/2) = 1/√n.
Choose bₙ = 1/√n = 1/n^(1/2). p = 1/2 ≤ 1 ⟹ Σ 1/√n diverges.
Compute L: L = lim [(√n + 1)/(n + 3)] / [1/√n] = lim √n · (√n + 1)/(n + 3) = lim (n + √n)/(n + 3) Divide by n: = lim (1 + 1/√n)/(1 + 3/n) = 1/1 = 1
L = 1, so 0 < L < ∞. Both series share fate. Σ 1/√n diverges ⟹ Σ (√n + 1)/(n + 3) diverges.
Diverges

Example 7 — LCT with a square root in the denominator

Does Σ_n=1^∞ 1/√(n² + n + 1) converge?

Dominant terms: √(n² + n + 1) ≈ √(n²) = n. So aₙ ≈ 1/n.
Choose bₙ = 1/n. Harmonic series Σ 1/n diverges.
Compute L: L = lim [1/√(n² + n + 1)] / [1/n] = lim n/√(n² + n + 1) Divide inside root by n²: = lim n / (n√(1 + 1/n + 1/n²)) = lim 1/√(1 + 1/n + 1/n²) = 1/√1 = 1
L = 1, so both series share fate. Σ 1/n diverges ⟹ Σ 1/√(n² + n + 1) diverges.
Diverges

Example 8 — LCT: mixed polynomial and exponential

Test Σ_n=1^∞ n/(2ⁿ + 3ⁿ) for convergence.

Dominant terms: For large n, 3ⁿ dominates 2ⁿ (since 3 > 2), so 2ⁿ + 3ⁿ ≈ 3ⁿ. Also n is polynomial, much smaller than 3ⁿ. So aₙ ≈ n/3ⁿ.
Choose bₙ = 1/2ⁿ = (1/2)ⁿ. Geometric series with |r| = 1/2 < 1 ⟹ converges. (n/3ⁿ also converges, but (1/2)ⁿ is simpler to work with.)
Compute L: L = lim [n/(2ⁿ + 3ⁿ)] / [1/2ⁿ] = lim n · 2ⁿ/(2ⁿ + 3ⁿ) Divide by 3ⁿ: = lim n · (2/3)ⁿ / ((2/3)ⁿ + 1) Since (2/3)ⁿ → 0 and n grows, the numerator → 0 and denominator → 1. L = 0.
L = 0. Since L = 0 and Σ (1/2)ⁿ converges, LCT (L=0 case) tells us Σaₙ also converges.
Converges

§ 05DCT vs LCT — When to Use Each

Direct Comparison Test

Best for: Series where the inequality is obvious: typically when you add a positive constant to the denominator, making it larger and the fraction smaller.

Requires: A provable term-by-term inequality for all n ≥ N.

Quick wins: 1/(n² + 5) ≤ 1/n², 1/(2ⁿ − 1) ≤ 2/2ⁿ for n ≥ 2.

Pitfall: Sometimes the inequality goes the wrong way: adding a positive term to the numerator makes things larger, which is useful for divergence but not convergence.

Limit Comparison Test

Best for: Rational-function-style series where you want to strip away lower-order terms. No need to prove a pointwise inequality.

Requires: The limit L = lim aₙ/bₙ to be finite and positive (or use the L = 0/∞ special cases carefully).

Quick wins: Any series of the form (polynomial)/(polynomial), or (polynomial)/(polynomial under root).

Pitfall: If L = 0 or L = ∞, you only get a one-directional result, you need to check carefully which series is the "larger" one.

Feature	Direct Comparison Test	Limit Comparison Test
What you need to prove	aₙ ≤ bₙ (or aₙ ≥ bₙ) for all n ≥ N	lim aₙ/bₙ = L, where 0 < L < ∞
Direction of comparison	Must match: below a convergent, above a divergent	Either direction — both conclusions come at once if L is finite
Ease of use	Can require careful algebraic manipulation	Usually faster — just take a limit
Handles L = 0 or L = ∞	N/A (uses inequalities directly)	Yes, with limited one-directional conclusions
Fails for	Series with oscillating signs or mixed terms	Series where the ratio limit is 0 or ∞ (inconclusive in full generality)
Typical series type	Exponentials with subtraction, sin/cos bounds	Rational, root-of-polynomial denominators

§ 06Strategy Flowchart & Interactive Term Comparison

Decision Flowchart — Which Test to Reach For

Given a series Σaₙ with aₙ > 0 — can you identify a dominant term bₙ?

Inequality is obvious

Try the Direct Comparison Test (DCT)

Does the inequality go the "right" way?
(aₙ ≤ bₙ with Σbₙ conv., or aₙ ≥ bₙ with Σbₙ div.)

✓ Apply DCT — conclude same behaviour

Inequality is unclear

Try the Limit Comparison Test (LCT)

Compute L = lim aₙ/bₙ. Is 0 < L < ∞?

✓ Apply LCT — both series share convergence/divergence

Interactive — visualise aₙ vs bₙ term by term

Series aₙ:

Terms shown: 20

Gold bars = aₙ. Teal dots = comparison bₙ. The ratio aₙ/bₙ converges to L shown in the legend.

Interactive — ratio aₙ/bₙ converging to L

The ratio aₙ/bₙ stabilises at L as n → ∞. When 0 < L < ∞, both series behave the same way.

§ 07Further Worked Examples

Example 9 — Choosing between DCT and LCT

Determine convergence of Σ_n=1^∞ (n² + cos n)/(n⁵ + 2).

Dominant terms: n²/n⁵ = 1/n³. Suspect convergence (p = 3 > 1).
LCT is preferable because the cos n term makes a pointwise inequality tricky (cos n oscillates).
Choose bₙ = 1/n³. Σ 1/n³ converges.
Compute L: L = lim [(n² + cos n)/(n⁵ + 2)] / (1/n³) = lim n³(n² + cos n)/(n⁵ + 2) = lim (n⁵ + n³ cos n)/(n⁵ + 2) Divide by n⁵: = lim (1 + cos(n)/n²)/(1 + 2/n⁵) = 1/1 = 1 (since cos(n)/n² → 0 as n → ∞)
L = 1 ∈ (0, ∞). LCT: Σaₙ shares fate with Σ 1/n³ ⟹ converges. Converges

Example 10 — When DCT inequality goes the wrong way

A student attempts to use DCT on Σ 1/(n² − 1) (n ≥ 2) by writing 1/(n² − 1) > 1/n². Explain why this attempt fails and give a correct argument.

Why it fails: The student has aₙ > bₙ where bₙ = 1/n² is the convergent comparison. DCT only gives convergence when aₙ ≤ bₙ. Having aₙ ≥ bₙ with Σbₙ convergent tells us nothing useful about Σaₙ.
Correct DCT approach: Since n² − 1 < n², we have 1/(n² − 1) > 1/n². This is the right direction for divergence, but we don't have a divergent comparison here. Switch to LCT.
LCT with bₙ = 1/n²: L = lim [1/(n² − 1)] / (1/n²) = lim n²/(n² − 1) = 1
L = 1. Both series share fate. Σ 1/n² converges ⟹ Σ 1/(n² − 1) converges.
Converges, but the DCT shortcut attempted by the student was invalid.

Example 11 — Series with a logarithm

Does Σ_n=2^∞ 1/(n ln n) converge?

First thought: 1/(n ln n) < 1/n and the harmonic series diverges: but this inequality goes the wrong way for DCT (smaller than a divergent series tells us nothing).
LCT with bₙ = 1/n: L = lim [1/(n ln n)] / (1/n) = lim 1/(ln n) = 0
L = 0 and Σ 1/n diverges. The L = 0 case of LCT only helps when Σbₙ converges: this gives no conclusion.
Use the Integral Test directly: ∫₂^∞ 1/(x ln x) dx = [ln(ln x)]₂^∞ → ∞. The integral diverges.
By the Integral Test: Σ 1/(n ln n) diverges. Diverges
Key lesson: When comparison tests fail (L = 0 or ∞, wrong direction), switch to the Integral Test or Ratio Test.

Example 12 — Three-step problem: identify, compare, conclude

Determine whether Σ_n=1^∞ (2n³ − n + 1)/(5n⁵ + 4n² + 7) converges.

Dominant terms: 2n³/5n⁵ = 2/(5n²) ~ 1/n². Expect convergence.
Choose bₙ = 1/n². p-series, p = 2 > 1, converges.
Compute L: L = lim [(2n³ − n + 1)/(5n⁵ + 4n² + 7)] · n² = lim (2n⁵ − n³ + n²)/(5n⁵ + 4n² + 7) Divide by n⁵: = lim (2 − 1/n² + 1/n³)/(5 + 4/n³ + 7/n⁵) = 2/5
L = 2/5 ∈ (0, ∞). LCT: Σaₙ and Σbₙ share fate. Σ 1/n² converges ⟹ Σaₙ converges.
Converges

§ 08Common Mistakes

Mistake 1 — Inequality in the wrong direction for DCT Showing aₙ ≤ bₙ where Σbₙ diverges tells you nothing. Showing aₙ ≥ bₙ where Σbₙ converges tells you nothing. The DCT only works when the known series is doing the "right" job: a convergent upper bound, or a divergent lower bound.

Mistake 2 — Applying LCT when L = 0 or L = ∞ and concluding both ways If L = lim aₙ/bₙ = 0, you may only conclude Σaₙ converges when Σbₙ converges. You cannot conclude Σaₙ diverges if Σbₙ diverges. Example: Σ 1/n² vs Σ 1/n gives L = 0. Σ 1/n diverges, but Σ 1/n² converges — the LCT (L=0 case) was correct to say nothing about this direction.

Mistake 3 — Choosing the wrong comparison bₙ Picking bₙ = 1/n for a series aₙ that actually behaves like 1/n² causes all the limits to go to 0 or ∞ — the test becomes inconclusive. Always identify the correct dominant term first by keeping only highest-degree numerator and denominator terms.

Mistake 4 — Forgetting positivity requirements Both comparison tests require aₙ ≥ 0 (and bₙ > 0 for LCT). For series with negative terms, you need the Alternating Series Test or Absolute Convergence first. Never apply DCT or LCT blindly to a series with mixed signs.

Mistake 5 — Not confirming Σbₙ's status before applying DCT You must state and justify whether your comparison series converges or diverges before invoking DCT. Writing "aₙ ≤ 1/n² therefore Σaₙ converges" skips the necessary step of confirming Σ 1/n² converges (by p-series, p=2 > 1).

§ 09Related Pages

Comparison tests sit at the heart of a larger family of convergence tools. The pages below are direct prerequisites or natural next steps.

Comparison Tests
for Infinite Series

§ 01The Big Idea — Learning from a Known Neighbour

§ 02The Comparison Arsenal — Series You Already Know

§ 03The Direct Comparison Test (DCT)

Step-by-Step Method for the DCT

Worked Examples — Direct Comparison Test

§ 04The Limit Comparison Test (LCT)

Why Does This Work?

Step-by-Step Method for the LCT

Worked Examples — Limit Comparison Test

§ 05DCT vs LCT — When to Use Each

§ 06Strategy Flowchart & Interactive Term Comparison

Decision Flowchart — Which Test to Reach For

§ 07Further Worked Examples

§ 08Common Mistakes

§ 09Related Pages

§ 10Quiz — Test Your Understanding