Tags: series, sequences, geometric series, divergence test, limit comparison test, alternating series, power series, Taylor polynomial, radius of convergence, interval of convergence, Calculus II, Texas A&M, MATH 172, MATH 2414
Series and sequences form the backbone of the second half of Calc II. You need to know how to evaluate geometric series, test for convergence (Divergence Test, Limit Comparison, Alternating Series Test), find radii and intervals of convergence for power series, and build Taylor polynomials. Most exam questions in this area test whether you can pick the right convergence test and apply it cleanly.
Sequence
An ordered list of numbers a₁, a₂, a₃, … defined by a formula aₙ. A sequence converges if limₙ→∞ aₙ exists and is finite.
Series
The sum of a sequence's terms: Σaₙ. A series converges if its partial sums approach a finite limit.
Geometric series
A series of the form Σ₍ₙ₌₀₎^∞ arⁿ. Converges to a/(1 − r) when |r| < 1; diverges otherwise.
p-series
A series of the form Σ1/nᵖ. Converges when p > 1; diverges when p ≤ 1.
Harmonic series
The p-series with p = 1: Σ1/n. It diverges.
Divergence Test (nth-term test)
If limₙ→∞ aₙ ≠ 0 (or the limit does not exist), then Σaₙ diverges. If the limit is zero, the test is inconclusive.
Limit Comparison Test
Given Σaₙ and Σbₙ with positive terms, if limₙ→∞ aₙ/bₙ = L where 0 < L < ∞, then both series converge or both diverge.
Alternating Series Test (Leibniz test)
Σ(−1)ⁿbₙ converges if bₙ is positive, decreasing, and limₙ→∞ bₙ = 0.
Absolute convergence
A series Σaₙ converges absolutely if Σ|aₙ| converges. Absolute convergence implies convergence.
Conditional convergence
A series converges conditionally if it converges but does not converge absolutely.
Power series
A series of the form Σcₙ(x − a)ⁿ. It converges on some interval centred at a.
Radius of convergence (R)
The distance from the centre a within which a power series converges. The series converges for |x − a| < R and diverges for |x − a| > R.
Interval of convergence
The set of all x values for which a power series converges, including any endpoint(s) where it converges.
Taylor polynomial
The nth-degree polynomial approximation of f(x) near x = a: Pₙ(x) = Σ₍ₖ₌₀₎ⁿ f⁽ᵏ⁾(a)/k! · (x − a)ᵏ.
The sum formula is Σ₍ₙ₌₀₎^∞ arⁿ = a/(1 − r), valid only when |r| < 1.
Identify the first term a and the common ratio r
If |r| ≥ 1, the series diverges
Worked example: Σ₍ₙ₌₀₎^∞ 5·(2/3)ⁿ
a = 5, r = 2/3
|2/3| < 1, so it converges
Sum = 5/(1 − 2/3) = 5/(1/3) = 15
If limₙ→∞ aₙ ≠ 0, the series Σaₙ diverges.
This test can only prove divergence, never convergence
A limit of zero tells you nothing on its own; you need a different test to confirm convergence
The classic trap: the harmonic series has limₙ→∞ 1/n = 0 but still diverges
Useful when aₙ "looks like" a simpler series bₙ for large n.
Worked example: Does Σ n/(3n² + 1) converge or diverge?
For large n, n/(3n² + 1) behaves like n/(3n²) = 1/(3n)
Compare with bₙ = 1/n
L = limₙ→∞ [n/(3n² + 1)] / [1/n] = limₙ→∞ n²/(3n² + 1) = 1/3
Since 0 < 1/3 < ∞ and Σ1/n diverges (harmonic), the original series also diverges
Worked example: Does Σ(−1)ⁿ/√n converge absolutely, conditionally, or neither?
Step 1, test absolute convergence:
Σ|aₙ| = Σ1/√n = Σ1/n^(1/2)
This is a p-series with p = 1/2 ≤ 1, so it diverges
Not absolutely convergent
Step 2, apply the Alternating Series Test:
bₙ = 1/√n is positive
bₙ is decreasing (since √n is increasing)
limₙ→∞ 1/√n = 0
All three conditions met, so the series converges
Conclusion: the series is conditionally convergent.
To find R, apply the Ratio Test to the general term.
Worked example: Find the interval of convergence for Σ (x − 3)ⁿ / (n · 2ⁿ)
Step 1, Ratio Test:
limₙ→∞ |aₙ₊₁/aₙ| = |x − 3|/2
Converges when |x − 3|/2 < 1, i.e. |x − 3| < 2
Radius R = 2, preliminary interval (1, 5)
Step 2, check x = 1:
Σ(−2)ⁿ/(n · 2ⁿ) = Σ(−1)ⁿ/n
This is the alternating harmonic series, which converges
Step 3, check x = 5:
Σ2ⁿ/(n · 2ⁿ) = Σ1/n
This is the harmonic series, which diverges
Interval of convergence: [1, 5).
Special case: Σ n!xⁿ
Ratio Test gives limₙ→∞ (n + 1)|x| = ∞ for any x ≠ 0
Converges only at x = 0
Radius of convergence R = 0
Worked example: Find limₙ→∞ ln(n)/n
This is ∞/∞ form
Apply L'Hopital's Rule to the continuous version f(x) = ln(x)/x
limₓ→∞ (1/x)/1 = 0
So the sequence converges to 0
The second-degree Taylor polynomial centred at a is:
P₂(x) = f(a) + f'(a)(x − a) + f''(a)/2! · (x − a)²
Worked example: P₂(x) for f(x) = √x centred at a = 4
f(x) = x^(1/2), so f(4) = 2
f'(x) = (1/2)x^(−1/2), so f'(4) = 1/4
f''(x) = −(1/4)x^(−3/2), so f''(4) = −1/32
P₂(x) = 2 + (1/4)(x − 4) − (1/64)(x − 4)²
Formula | Expression |
|---|---|
Geometric series sum | a/(1 − r), |r| < 1 |
p-series convergence | Σ1/nᵖ converges iff p > 1 |
Ratio Test limit | limₙ→∞ |aₙ₊₁/aₙ| < 1 → converges |
Taylor polynomial | Pₙ(x) = Σ f⁽ᵏ⁾(a)/k! · (x − a)ᵏ |
⚠️ The Divergence Test can only prove divergence. If limₙ→∞ aₙ = 0, you have proved nothing. This is the most common misapplication on exams.
⚠️ When using the Limit Comparison Test, your comparison series bₙ must have known convergence behaviour. The harmonic series (diverges) and p-series (converges for p > 1) are the go-to comparisons.
⚠️ Always check endpoints when finding an interval of convergence. The Ratio Test is inconclusive at |x − a| = R, so you must test each endpoint with a separate convergence test.
⚠️ For Taylor polynomials, be careful with the factorial in the denominator. The coefficient of (x − a)² is f''(a)/2!, not f''(a).
⚠️ "Conditionally convergent" means it converges but not absolutely. You must show both parts: that Σ|aₙ| diverges and that the original series converges.
Q: Find the sum of Σ₍ₙ₌₀₎^∞ 5·(2/3)ⁿ.
A: 15. Use the geometric series formula: 5/(1 − 2/3) = 15.
Q: Does the Divergence Test say that if limₙ→∞ aₙ = 0, the series converges?
A: No. If the limit is zero, the Divergence Test is inconclusive. You need a different test.
Q: Use the Limit Comparison Test with bₙ = 1/n to determine whether Σ n/(3n² + 1) converges or diverges.
A: Diverges. The limit aₙ/bₙ = 1/3 (positive, finite), and Σ1/n diverges, so the original series diverges.
Q: Is Σ(−1)ⁿ/√n absolutely convergent, conditionally convergent, or divergent?
A: Conditionally convergent. Σ1/√n diverges (p = 1/2), but the Alternating Series Test confirms convergence.
Q: What is the interval of convergence for Σ (x − 3)ⁿ/(n · 2ⁿ)?
A: [1, 5). Ratio Test gives R = 2. At x = 1, alternating harmonic (converges). At x = 5, harmonic (diverges).
Q: What is the radius of convergence of Σ n!xⁿ?
A: R = 0. The series converges only at x = 0.
Q: Find limₙ→∞ ln(n)/n.
A: 0. Apply L'Hopital's Rule to ln(x)/x.
Q: Find the second-degree Taylor polynomial for √x centred at a = 4.
A: P₂(x) = 2 + (1/4)(x − 4) − (1/64)(x − 4)².
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