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How do you test if a series converges?
Compute limn→∞ an
if limit = finite number, series converges
if limit = ±∞ or DNE, series diverges
Squeeze Theorem for Sequences
if:
an ≤ cn ≤ bn
and:
lim an = lim bn = L
then:
lim cn = L
Definition of a series…
∑an = limn→∞ ∑nk = 1 ak
you are checking for partial sums
Geometric Series Test
Form: ∑arn or ∑arn-1
Condition: | r | < 1
Result: ∑arn = a / (1 - r)
if: | r | ≥ 1, then divergent
Divergence Test (nth-term test)
check: limn→∞ an
if limit ≠ 0, then diverges
if limit = 0, inconclusive
Integral Test (When do you use it?)
You need ALL 3:
f(x) = an
positive, continuous, decreasing
Integral Test - What does it actually mean and how do I use it?
Step 1 (Pattern Recognition): ∑an = ∑ f(n)
Step 2 (Turn it into an Integral) ∫∞1 f(x) dx
Step 3:
If the integral gives a finite number = convergent
if the integral = ∞ (blows up) = divergent
Direct Comparison Test — When to use?
Use When:
Terms look similar to known series
Usally compare with, 1 / (np) or geometric.
Direct Comparison — Conditions
for an, bn > 0…
if: an ≤ bn
and ∑bn converges → ∑an converges
if: an ≥ bn
and ∑bn diverges → ∑an diverges
Limit Comparison Test — When to Use?
Use when:
Fractions with similar powers
hard to compare directly
Limit Comparison — Result
Compute: limn→∞ (an/bn) = L
if 0 < L < ∞ → SAME Behavior (Both converge or Both diverge)
Alternating Series Test (AST) — Conditions
Series: ∑(-1)nbn or (-1)n+1bn
You need: bn decreasing and lim bn = 0
AST - What does it Prove?
Series Converges (but only conditionally)
Absolute vs Conditional Convergence
if….
∑|an| converges, then absolutely convergent
if…
∑an converges but ∑|an| diverges, then conditionally convergent
Ratio Test — When to use?
Best for:
Factorials
Exponentials
Powers like nn
Ratio Test — formula
L = limn→∞ |(an+1/an)|
Ratio Test — Result
L < 1 → Converges (Absolutely)
L > 1 → Diverges
L = 1 → Inconclusive
Root Test — When to use
Best for:
powers like (an)n
Root Test — Formula
L = limn→∞ n√|an|
Root Test — Result
L < 1 → Converges (Absolutely)
L > 1 → Diverges
L = 1 → Inconclusive
p-Series Test
∑(1/np)
p > 1 Converges
p ≤ 1 → Diverges
Harmonic Series
∑1/n → Diverges
Divergence Test Formula
lim n→∞ an ≠ 0 → Diverges
Ratio Test Formula
L = limn→∞ |(an+1/an)|
Root Test Formula
L = limn→∞ n√|an|
Limit Comparison Formula
limn→∞ (an/bn) = L
Integral Test Setup
∑an ↔ ∫∞1 f(x) dx
Alternating Series Form
∑(-1)n bn or (-1)n+1 bn