Every major test for infinite series convergence — geometric series, p-series, divergence test, integral test, comparison, ratio, root, and alternating series — with clear criteria, worked examples, and a decision guide.
An infinite series is the sum of infinitely many terms: ∑ aₙ = a₁ + a₂ + a₃ + ···. The series converges if the sequence of partial sums Sₙ = a₁ + a₂ + ··· + aₙ approaches a finite limit L as n → ∞. Otherwise, it diverges.
Partial Sum
Sₙ = a₁ + a₂ + ··· + aₙ
The sum of the first n terms. The series converges if lim(n→∞) Sₙ = L (finite).
Key Principle
The terms aₙ → 0 is necessary but not sufficient for convergence. The harmonic series ∑ 1/n has terms → 0 yet diverges. A test beyond the divergence test is always needed to confirm convergence.
The geometric series is the most important series in calculus. Every other test is ultimately benchmarked against it.
General Form
Convergence Condition
The series converges if and only if the common ratio r satisfies |r| < 1. If |r| ≥ 1, the series diverges.
Sum Formula
When |r| < 1 and the series starts at n = 0, the exact sum is a/(1 − r), where a is the first term (n = 0 term).
p-Series Form
Converges: p > 1
Diverges: p ≤ 1
The p-series is the go-to benchmark for the comparison and limit comparison tests. When a series looks like 1/n^p after simplification, apply the p-series test directly.
Critical Warning
If lim aₙ = 0, the test is inconclusive — the series may still diverge. The harmonic series ∑ 1/n has aₙ → 0 yet diverges. The divergence test can only prove divergence, never convergence.
Best Practice
Always apply the divergence test first. If the limit isn't zero, you stop immediately. It takes 30 seconds and can save you from a lengthy integral or ratio test computation.
Statement: Let f be continuous, positive, and decreasing on [1, ∞) with f(n) = aₙ. Then:
The test doesn't compute the sum — only determines convergence. The integral and series may converge to different values.
Requirements Checklist
Worked Example: ∑ 1/(n ln n)
f(x) = 1/(x ln x) — continuous, positive, decreasing on [2, ∞)
∫₂^∞ 1/(x ln x) dx = [ln(ln x)]₂^∞ = ∞
Integral diverges → series ∑ 1/(n ln n) diverges
Worked Example: ∑ 1/(n² + 1)
f(x) = 1/(x² + 1) — continuous, positive, decreasing on [1, ∞)
∫₁^∞ 1/(x² + 1) dx = [arctan x]₁^∞ = π/2 − π/4 = π/4
Integral converges → series ∑ 1/(n² + 1) converges
Requires 0 ≤ aₙ ≤ bₙ for all sufficiently large n:
∑ bₙ converges → ∑ aₙ converges
A smaller series squeezed under a convergent one must also converge.
∑ aₙ diverges → ∑ bₙ diverges
A larger series sitting above a divergent one must also diverge.
Example: ∑ 1/(n² + 5)
Compare with ∑ 1/n² (p-series, p=2 → converges)
Since 1/(n² + 5) ≤ 1/n² for all n ≥ 1:
∑ 1/(n² + 5) converges by direct comparison
If aₙ, bₙ > 0 and lim(n→∞) aₙ/bₙ = L with 0 < L < ∞, then ∑ aₙ and ∑ bₙ either both converge or both diverge.
L ∈ (0, ∞)
Both series behave the same — same fate.
L = 0
∑ bₙ converges → ∑ aₙ converges.
L = ∞
∑ bₙ diverges → ∑ aₙ diverges.
Example: ∑ (3n + 1) / (n³ − 2)
Compare with bₙ = 3/n² (p-series, p=2 → converges)
lim [(3n + 1)/(n³ − 2)] / [3/n²] = lim (3n³ + n²) / (3n³ − 6) = 1
L = 1 ∈ (0, ∞) and ∑ 3/n² converges
∑ (3n + 1)/(n³ − 2) converges by limit comparison
Compute the limit
L < 1
Series converges absolutely
L > 1 or L = ∞
Series diverges
L = 1
Inconclusive — try another test
Best Used When
The general term involves factorials (n!), exponentials (a^n), or products of polynomials and exponentials — because these simplify cleanly when you form aₙ₊₁/aₙ. The ratio test is one of the most powerful tests in practice.
Worked Examples
∑ n! / n^n
|aₙ₊₁/aₙ| = [(n+1)! / (n+1)^(n+1)] · [n^n / n!]
= (n+1) · n^n / (n+1)^(n+1) = n^n / (n+1)^n = (n/(n+1))^n
lim (n/(n+1))^n = lim 1/(1+1/n)^n = 1/e < 1
Converges (L = 1/e < 1)
∑ 2^n / n!
|aₙ₊₁/aₙ| = 2^(n+1)/(n+1)! · n!/2^n = 2/(n+1)
lim 2/(n+1) = 0 < 1
Converges (L = 0 < 1)
∑ n^n / n!
|aₙ₊₁/aₙ| = (n+1)^(n+1)/(n+1)! · n!/n^n = (1 + 1/n)^n → e
L = e > 1
Diverges (L = e > 1)
Compute the limit
L < 1
Converges absolutely
L > 1 or L = ∞
Diverges
L = 1
Inconclusive
Best Used When
The general term is raised to the nth power — written as (f(n))^n — because the nth root cancels the exponent cleanly. If you see aₙ = (something)^n, try the root test before the ratio test.
Worked Examples
∑ (2n/(n+3))^n
|aₙ|^(1/n) = 2n/(n+3)
lim 2n/(n+3) = 2 > 1
Diverges (L = 2 > 1)
∑ (n/(2n+1))^n
|aₙ|^(1/n) = n/(2n+1)
lim n/(2n+1) = 1/2 < 1
Converges (L = 1/2 < 1)
An alternating series has terms that alternate in sign: ∑ (−1)^n bₙ or ∑ (−1)^(n+1) bₙ, where bₙ > 0.
Convergence Conditions (both must hold)
bₙ is eventually decreasing
bₙ₊₁ ≤ bₙ for all sufficiently large n
lim(n→∞) bₙ = 0
The terms must approach zero
Alternating Harmonic Series
∑ (−1)^(n+1)/n
bₙ = 1/n is decreasing and → 0. Both conditions met → converges. Sum = ln 2 ≈ 0.693. Note: ∑ 1/n diverges, so this is conditionally convergent.
Alternating p-Series
∑ (−1)^n / n²
bₙ = 1/n² is decreasing and → 0. Both conditions met → converges. Also ∑ 1/n² converges (p=2), so this is absolutely convergent.
Alternating Series Remainder Estimate
If Sₙ is the nth partial sum of a convergent alternating series satisfying the Leibniz conditions, then the error |S − Sₙ| ≤ bₙ₊₁ (the absolute value of the first omitted term). This gives a simple error bound for approximations.
If the series of absolute values converges, the original series converges absolutely. Absolute convergence implies convergence. Rearranging the terms of an absolutely convergent series does not change its sum.
Example: ∑ (−1)^n/n² converges absolutely since ∑ 1/n² converges.
The series converges only due to cancellation of positive and negative terms. By the Riemann rearrangement theorem, a conditionally convergent series can be rearranged to converge to any value or to diverge.
Example: ∑ (−1)^(n+1)/n converges to ln 2, but ∑ 1/n diverges.
Testing Strategy
To test any series for absolute convergence, apply a convergence test to ∑ |aₙ|. If that converges, you have absolute convergence. If ∑ |aₙ| diverges but ∑ aₙ converges (e.g., by the alternating series test), you have conditional convergence.
| Test | Apply When | Converges If | Limitation |
|---|---|---|---|
| Divergence Test | Always try first | Cannot prove convergence | Only proves divergence |
| Geometric Series | ∑ ar^n form | |r| < 1 | Must be exactly geometric |
| p-Series | ∑ 1/n^p form | p > 1 | Must be exactly p-series |
| Integral Test | f(x) continuous, positive, decreasing | ∫₁^∞ f dx converges | Doesn't give the sum |
| Comparison | Can bound aₙ above/below | aₙ ≤ bₙ and ∑ bₙ converges | Need to find the right benchmark |
| Limit Comparison | Terms behave like a known series | lim aₙ/bₙ = L ∈ (0, ∞), ∑ bₙ same fate | Requires positive terms |
| Ratio Test | Factorials or exponentials in aₙ | L = lim |aₙ₊₁/aₙ| < 1 | Inconclusive if L = 1 |
| Root Test | Term raised to nth power | L = lim |aₙ|^(1/n) < 1 | Inconclusive if L = 1 |
| Alternating Series | Signs alternate, bₙ > 0 | bₙ decreasing and bₙ → 0 | Only for alternating series |
Does lim(n→∞) aₙ ≠ 0?
Yes → Diverges by the divergence test. Done.
Is it exactly ∑ ar^n?
Yes → Geometric series test: converges iff |r| < 1.
Is it exactly ∑ 1/n^p?
Yes → p-series test: converges iff p > 1.
Does the term contain n! or a^n?
Yes → Try the ratio test. Compute lim |aₙ₊₁/aₙ|.
Is the term of the form (f(n))^n?
Yes → Try the root test. Compute lim |aₙ|^(1/n).
Do the signs alternate?
Yes → Try the alternating series test (Leibniz). Check bₙ decreasing and bₙ → 0.
Does the term look like a simpler series?
Yes → Try limit comparison with the dominant behavior (drop lower-order terms).
Can you compute ∫₁^∞ f(x) dx?
Yes → Use the integral test (confirm f is continuous, positive, decreasing).
Strategy: factorial and exponential → ratio test
aₙ₊₁/aₙ = [3^(n+1)·(n+1)! / (n+1)^(n+1)] · [n^n / (3^n · n!)]
= 3(n+1) · n^n / (n+1)^(n+1)
= 3 · n^n / (n+1)^n
= 3 · (n/(n+1))^n = 3 · (1/(1+1/n))^n → 3/e
3/e ≈ 3/2.718 ≈ 1.104 > 1
Diverges (L = 3/e > 1)
Alternating series. Let bₙ = 1/(n² + n + 1).
Step 1: bₙ → 0 as n → ∞ ✓
Step 2: bₙ is decreasing (denominator increases) ✓
Converges by the alternating series test
Absolute convergence check: ∑ 1/(n² + n + 1) behaves like ∑ 1/n² (p=2 > 1)
Also absolutely convergent
Strategy: exponential → ratio test
|aₙ₊₁/aₙ| = [(n+1)/2^(n+1)] · [2^n/n]
= (n+1)/(2n)
lim (n+1)/(2n) = 1/2 < 1
Converges absolutely (L = 1/2 < 1)
Strategy: integral test (no factorial, no exponential, not obviously a p-series)
f(x) = 1/(x · √(ln x)), continuous, positive, decreasing on [2, ∞)
Let u = ln x, du = dx/x:
∫ dx/(x · √(ln x)) = ∫ u^(−1/2) du = 2√u = 2√(ln x)
∫₂^∞ f dx = [2√(ln x)]₂^∞ = ∞
Integral diverges → series diverges
Strategy: term raised to nth power → root test
|aₙ|^(1/n) = [(1 + 1/n)^(n²)]^(1/n) = (1 + 1/n)^n
lim (1 + 1/n)^n = e > 1
Diverges (L = e > 1)
A geometric series ∑ ar^n converges if and only if |r| < 1, where r is the common ratio. When it converges, the sum is S = a / (1 − r), where a is the first term. If |r| ≥ 1, the series diverges. Example: ∑ (1/3)^n has r = 1/3, so |r| < 1 and it converges. Starting from n=0, S = 1 / (1 − 1/3) = 3/2.
The p-series ∑ 1/n^p converges if p > 1 and diverges if p ≤ 1. The harmonic series ∑ 1/n has p = 1 and diverges, even though its terms approach 0. The series ∑ 1/n² has p = 2 > 1 and converges to π²/6. The p-series test is often used as a comparison benchmark for other series.
The divergence test states: if lim(n→∞) aₙ ≠ 0, then ∑ aₙ diverges. It can only prove divergence — it cannot prove convergence. If lim aₙ = 0, the series may still diverge (the harmonic series is the classic example). Always try the divergence test first since it's quick; if the limit isn't zero, you're done.
Use the ratio test when the general term involves factorials (n!) or exponentials like a^n, since factorials simplify nicely in the ratio aₙ₊₁/aₙ. Use the root test when the general term is raised to the nth power, like (f(n))^n, since the nth root eliminates the exponent cleanly. If L = lim |aₙ₊₁/aₙ| or L = lim |aₙ|^(1/n): L < 1 means converges absolutely, L > 1 means diverges, L = 1 is inconclusive.
The alternating series test (Leibniz's theorem) applies to series of the form ∑ (−1)^n bₙ where bₙ > 0. Two conditions must hold: (1) the terms must be eventually decreasing — bₙ₊₁ ≤ bₙ for all large n, and (2) lim(n→∞) bₙ = 0. If both conditions are met, the series converges. The error in approximating the sum with the first N terms is bounded by |bₙ₊₁|, the first omitted term.
A series ∑ aₙ converges absolutely if ∑ |aₙ| converges. It converges conditionally if ∑ aₙ converges but ∑ |aₙ| diverges. Absolute convergence is stronger: if a series converges absolutely, it also converges. The alternating harmonic series ∑ (−1)^(n+1)/n converges (to ln 2) but ∑ 1/n diverges, so it is conditionally convergent. By the Riemann rearrangement theorem, a conditionally convergent series can be rearranged to sum to any value.
The integral test links series to improper integrals. If f(x) is continuous, positive, and decreasing on [1, ∞), and aₙ = f(n), then ∑ aₙ and ∫₁^∞ f(x) dx either both converge or both diverge. The test doesn't give the sum of the series — only whether it converges. It works best when you can actually compute the improper integral. Example: ∑ 1/n² converges because ∫₁^∞ 1/x² dx = 1 converges.
Follow this decision flow: (1) Check if lim aₙ ≠ 0 — if so, diverges by the divergence test. (2) If it's a geometric series ∑ ar^n, use the geometric series test. (3) If it's ∑ 1/n^p, use the p-series test. (4) If the term has n! or a^n, try the ratio test. (5) If the term is raised to the nth power like (f(n))^n, try the root test. (6) If it's an alternating series, try the alternating series test. (7) If the term looks like a known convergent/divergent series, try comparison or limit comparison.
The direct comparison test: if 0 ≤ aₙ ≤ bₙ for all large n, then convergence of ∑ bₙ implies convergence of ∑ aₙ, and divergence of ∑ aₙ implies divergence of ∑ bₙ. The limit comparison test is easier to apply when terms differ by complicated expressions: if aₙ, bₙ > 0 and lim(n→∞) aₙ/bₙ = L where 0 < L < ∞, then ∑ aₙ and ∑ bₙ either both converge or both diverge. The limit comparison test avoids the need for strict inequalities.
Radius and interval of convergence, Taylor and Maclaurin series, term-by-term differentiation and integration
Monotone sequences, bounded sequences, squeeze theorem, limits of sequences, and convergence criteria
Integrals over infinite intervals, discontinuous integrands, comparison test for integrals
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