Series Convergence Tests

What "convergence" actually means

Given a sequence a₁, a₂, a₃, ... define the partial sums:

S_N = a₁ + a₂ + ... + a_N

The series ∑_{n=1}^∞ aₙ converges if the sequence S_N has a finite limit S; the limit is then declared the "sum" of the series. Otherwise the series diverges. There's no other definition.

The simplest necessary condition: if aₙ does not tend to zero, the series cannot converge. In symbols, ∑ aₙ converges ⇒ aₙ → 0. The converse fails — the harmonic series ∑ 1/n diverges even though 1/n → 0 — so aₙ → 0 is necessary but not sufficient. This one-way implication is the divergence test: if aₙ does not approach 0, you're done, the series diverges.

Six tests at a glance

Test	When to use	Conclusion	Limitations
Integral test	aₙ = f(n), f positive, decreasing, continuous	∑ aₙ and ∫₁^∞ f(x) dx converge or diverge together	Need f monotone and to evaluate the integral
Comparison test	0 ≤ aₙ ≤ bₙ for known bₙ	∑ bₙ converges ⇒ ∑ aₙ converges; ∑ aₙ diverges ⇒ ∑ bₙ diverges	Requires explicit term-by-term inequality
Limit comparison	aₙ > 0, bₙ > 0, lim aₙ/bₙ = c, finite, positive	Both converge or both diverge	Requires a good benchmark series bₙ
Ratio test	Factorials, powers like rⁿ, n^k	L = lim |a_{n+1}/aₙ|: L < 1 converges absolutely; L > 1 diverges; L = 1 inconclusive	Inconclusive at L = 1; especially weak for rational aₙ
Root test	Powers of n, terms like nⁿ or 2ⁿ	L = lim ⁿ√|aₙ|: L < 1 converges; L > 1 diverges; L = 1 inconclusive	Inconclusive at L = 1; sometimes harder to compute than ratio
Alternating series test	aₙ = (−1)ⁿ bₙ with bₙ ↘ 0	The series converges; |error| ≤ first omitted term	Only proves convergence, not absolute convergence
p-series test	∑ 1/nᵖ	Converges iff p > 1	Specific to that exact form

The integral test, ratio test, and root test all derive from comparison with a geometric or p-series; they're packaged shortcuts for the comparison test in their natural domains.

Integral test — calculus benchmarks

If f is positive, decreasing, and continuous on [1, ∞), and aₙ = f(n), then:

∑_{n=1}^∞ aₙ converges  ⇔  ∫₁^∞ f(x) dx converges

The reason is geometric: rectangles of width 1 and height f(n) under the curve sandwich the integral. Apply this to the harmonic series:

∫₁^∞ dx/x = lim_{b→∞} [ln x]₁^b = ∞    ⇒    ∑ 1/n diverges

And to the p-series:

∫₁^∞ dx/xᵖ = lim_{b→∞} [x^(1−p)/(1−p)] = finite iff p > 1

So ∑ 1/nᵖ converges iff p > 1. This single result — the p-series test — is the benchmark every other comparison test leans on.

Comparison and limit-comparison

The plain comparison test: if 0 ≤ aₙ ≤ bₙ and ∑ bₙ converges, then ∑ aₙ converges. Conversely, if ∑ aₙ diverges then ∑ bₙ diverges. The trick is finding the right benchmark — usually a p-series.

Limit comparison softens the requirement. If aₙ, bₙ > 0 and lim aₙ/bₙ = c with 0 < c < ∞, then ∑ aₙ and ∑ bₙ share a fate. Example: study ∑ (3n + 2) / (n² + 5n + 1). For large n the term behaves like 3/n, so compare to bₙ = 1/n:

aₙ/bₙ = (3n + 2)·n / (n² + 5n + 1) → 3 as n → ∞

The limit is finite and positive. Since ∑ 1/n diverges, so does the original series.

Ratio and root tests

Both tests find the asymptotic geometric ratio of the terms. The ratio test:

L = lim_{n→∞} |a_{n+1} / aₙ|
L < 1 ⇒ ∑ aₙ converges absolutely
L > 1 ⇒ ∑ aₙ diverges
L = 1 ⇒ inconclusive

The root test:

L = lim_{n→∞} ⁿ√|aₙ|
L < 1 ⇒ converges absolutely
L > 1 ⇒ diverges
L = 1 ⇒ inconclusive

Worked example with the ratio test — the series ∑ n!/nⁿ:

a_{n+1}/aₙ = ((n+1)! / (n+1)^(n+1)) · (nⁿ / n!)
           = (n+1) · nⁿ / (n+1)^(n+1)
           = (n / (n+1))ⁿ
           = 1 / (1 + 1/n)ⁿ → 1/e ≈ 0.368  < 1

L = 1/e < 1, so ∑ n!/nⁿ converges absolutely. The factorials win against nⁿ — surprisingly, since both grow rapidly. The ratio test is at its strongest when the terms involve factorials, exponentials, or n^k powers.

The root test handles cases the ratio test doesn't, like ∑ (n/(n+1))ⁿ². For each fixed n, ⁿ√aₙ = (n/(n+1))ⁿ → 1/e < 1, so the series converges. The ratio of consecutive terms in this series is messier; root test is cleaner.

Alternating series

An alternating series has terms with alternating signs: ∑ (−1)ⁿ bₙ where bₙ > 0. The alternating series test (Leibniz) says: if bₙ is monotonically decreasing and bₙ → 0, then ∑ (−1)ⁿ bₙ converges, and the truncation error is bounded by the first omitted term.

Apply this to ∑ (−1)ⁿ⁺¹/n = 1 − 1/2 + 1/3 − 1/4 + .... The bₙ = 1/n decreases to zero, so the series converges. Its sum is ln 2. Yet the same series with absolute values, ∑ 1/n, diverges. This is conditional convergence.

Absolute vs conditional convergence

	Absolutely convergent	Conditionally convergent
Definition	∑ |aₙ| converges	∑ aₙ converges, ∑ |aₙ| diverges
Convergence guaranteed?	Yes (absolute ⇒ ordinary)	Yes
Sum invariant under reordering?	Yes (Riemann)	No
Sum stable under absolute-value swaps?	Yes	Different rearrangements give different sums
Example	∑ (−1)ⁿ/n²	∑ (−1)ⁿ⁺¹/n = ln 2
Cauchy product with itself	Converges	May diverge

Riemann's rearrangement theorem makes conditional convergence pathological: by reordering the terms of a conditionally convergent series, you can make the sum equal any real number, +∞, or −∞. Absolute convergence is the safe regime where the sum is independent of ordering — essential for power series and most working theorems.

Picking a test in practice

1. Check the divergence test first. If aₙ ↛ 0, the series diverges and you're done.
2. Look at the form of aₙ:
   • Pure 1/nᵖ → p-series test.
   • Geometric / r^n term → geometric series formula.
   • Factorials or growing exponentials → ratio test.
   • Exponentials with n in the base or exponent → root test.
   • Rational function of n → limit comparison with a p-series.
   • Alternating signs and decreasing magnitude → alternating series test (and check absolute convergence separately if needed).
   • Hard, monotone, integrable form → integral test.
3. If your test is inconclusive (L = 1 in ratio/root), drop down to integral or comparison.

Where convergence tests matter

• Taylor expansions. The Taylor series of e^x converges for all x; sin and cos similarly. The series for ln(1 + x) converges only for |x| ≤ 1 (and at x = 1 conditionally). Determining the radius of convergence by the ratio test is standard practice; without it you don't know where the series represents the function.
• Fourier series. The convergence of Fourier coefficients dictates whether the series converges pointwise, uniformly, or only in L² — three different statements with three different consequences for whether you can differentiate the series term-by-term. Convergence tests on the coefficients separate the cases.
• Special functions. The Riemann zeta function ζ(s) = ∑ 1/nˢ converges only for Re(s) > 1 (the p-series test for complex p). Analytic continuation extends it to other s, but the original definition lives in that half-plane. Many physics calculations of vacuum energy and partition functions are founded on this.
• Generating functions. Combinatorial generating functions ∑ aₙ xⁿ have a radius of convergence that bounds the growth of aₙ (Cauchy-Hadamard formula). Whether the radius is positive, infinite, or zero tells you whether the sequence grows polynomially, slower than any exponential, or super-exponentially.

Common mistakes

• Concluding convergence from aₙ → 0. The condition is necessary, not sufficient. The harmonic series gives the classical counterexample: 1/n → 0, yet the sum diverges.
• Misapplying the ratio test at L = 1. You cannot conclude anything in this case. ∑ 1/n and ∑ 1/n² both give L = 1, with opposite verdicts. Switch to integral test or comparison.
• Forgetting the monotonicity hypothesis in the alternating-series test. The bₙ must be eventually decreasing. ∑ (−1)ⁿ aₙ with aₙ that oscillate or even grow can fail despite being alternating.
• Rearranging conditionally convergent series. Switching the order of terms in ∑ (−1)ⁿ⁺¹/n can change the sum (Riemann's theorem). Only absolute convergence licenses rearrangement.
• Comparing ∑ aₙ with ∑ bₙ when aₙ > bₙ for divergence. The plain comparison test runs only one way — bigger-than-divergent diverges, smaller-than-convergent converges. The opposite inequalities give no information.