Source: MATH 308, Texas A&M University, Sections 11.8–11.9
Tags: power series, radius of convergence, interval of convergence, geometric series, term-by-term differentiation, term-by-term integration, Maclaurin expansion, representing functions, ln(1−x), arctan, 1/(1−x)
A power series is an "infinite polynomial" ∑cₙ(x − a)ⁿ whose convergence depends on which x you plug in. The ratio test finds the radius of convergence R, and you check the endpoints separately. Once you know the geometric series 1/(1 − x) = ∑xⁿ for |x| < 1, you can derive power series for a wide family of functions by substitution, differentiation, and integration.
Power series (centred at a)
A series of the form ∑(n=0 to ∞) cₙ(x − a)ⁿ = c₀ + c₁(x − a) + c₂(x − a)² + … where the cₙ are constants, a is the centre, and x is a variable. When a = 0 this simplifies to ∑cₙxⁿ.
Radius of convergence (R)
The value R ≥ 0 (possibly ∞) such that the power series converges for |x − a| < R and diverges for |x − a| > R. Found using the ratio test (or root test).
Interval of convergence (I)
The set of all x for which the power series converges. Always an interval centred at a. Could be (a − R, a + R), [a − R, a + R], or half-open, depending on what happens at the endpoints. Also includes the degenerate cases I = {a} when R = 0 and I = (−∞, ∞) when R = ∞.
Geometric series (as a power series)
1/(1 − x) = ∑(n=0 to ∞) xⁿ = 1 + x + x² + x³ + … for |x| < 1. This is the starting point for most power-series representations in the course.
Term-by-term differentiation
If f(x) = ∑cₙ(x − a)ⁿ has radius R > 0, then f′(x) = ∑n·cₙ(x − a)^(n−1) for |x − a| < R. You differentiate each term as you would a polynomial.
Term-by-term integration
Under the same conditions, ∫f(x)dx = C + ∑cₙ(x − a)^(n+1)/(n + 1). The radius of convergence stays the same, though the interval of convergence may change at the endpoints.
A power series ∑cₙ(x − a)ⁿ is a function of x. For each particular x you substitute in, you get an ordinary numerical series that either converges or diverges.
Three things can happen:
The series converges for all x. Then R = ∞ and I = (−∞, ∞). Example: ∑xⁿ/n! (which equals eˣ).
The series converges only at x = a. Then R = 0 and I = {a}. Example: ∑n!·xⁿ.
The series converges for |x − a| < R and diverges for |x − a| > R for some finite R > 0. Endpoints need separate checking.
Apply the ratio test to the general term (treating x as a parameter):
lim |aₙ₊₁/aₙ| = |x − a| · (something depending on cₙ)
Set this less than 1 and solve for |x − a| to get R.
Example: ∑xⁿ. Ratio gives |x| < 1, so R = 1. At x = 1: ∑1 diverges. At x = −1: ∑(−1)ⁿ diverges. So I = (−1, 1).
Example: ∑(2ⁿ/n²)xⁿ. Ratio gives |aₙ₊₁/aₙ| = 2|x|·(n/(n+1))² → 2|x|. Converges when 2|x| < 1, i.e. |x| < 1/2, so R = 1/2. Check endpoints separately.
Example: ∑xⁿ/n!. Ratio gives |x|/(n+1) → 0 for every x. So R = ∞ and the series converges for all x.
At each endpoint, substitute the value of x and test the resulting numerical series using whatever applies: p-series, AST, divergence test, comparison, etc. The ratio test itself is always inconclusive at the endpoints (it gives L = 1 there).
Example: ∑(−1)ⁿ(x − 2)ⁿ/n. Ratio test gives |x − 2| < 1, so R = 1, centre a = 2.
At x = 3: ∑(−1)ⁿ · 1/n = alternating harmonic, converges.
At x = 1: ∑(−1)ⁿ(−1)ⁿ/n = ∑1/n, diverges.
Interval of convergence: (1, 3].
Example: ∑(−1)ⁿ(x − 1)ⁿ/3^(2n). Rewrite as ∑(−1)ⁿ(x − 1)ⁿ/9ⁿ = ∑(−(x−1)/9)ⁿ, a geometric series. Converges when |x − 1|/9 < 1, i.e. |x − 1| < 9. R = 9, I = (−8, 10). Endpoints give geometric series with |r| = 1, which diverge.
If you know that ∑cₙ(−2)ⁿ converges and ∑cₙ8ⁿ diverges, the radius of convergence satisfies 2 ≤ R ≤ 8. Convergence is guaranteed for |x| < 2 (inside the known convergent point), and divergence is guaranteed for |x| > 8. Values between 2 and 8 are inconclusive without more information. (Example 11.8.9 and 11.8.10 in the notes.)
The single identity
1/(1 − x) = ∑(n=0 to ∞) xⁿ, |x| < 1
generates most of the power series you need in this course by substitution, algebra, differentiation, or integration.
1/(1 + x²): Replace x with −x² in the geometric series. Result: ∑(−1)ⁿ x^(2n), |x| < 1.
1/(x + 3) around a = 0: Write as (1/3) · 1/(1 + x/3) = (1/3) · ∑(−x/3)ⁿ = ∑(−1)ⁿ xⁿ / 3^(n+1), |x| < 3.
x⁵/(x + 3): Multiply the series for 1/(x + 3) by x⁵. Each term picks up an extra x⁵.
Differentiating 1/(1 − x) = ∑xⁿ term by term gives:
1/(1 − x)² = ∑n·x^(n−1), |x| < 1
More generally, 1/(3 − x)² can be obtained by differentiating the series for 1/(3 − x).
Integrating 1/(1 − x) = ∑xⁿ term by term gives:
−ln(1 − x) = ∑xⁿ⁺¹/(n + 1) + C
Setting x = 0 gives C = 0, so:
ln(1 − x) = −∑(n=1 to ∞) xⁿ/n, −1 ≤ x < 1
Note the interval: the integrated series converges at x = −1 (by AST, giving ln 2), even though the original geometric series diverges there. Integration can expand the interval at an endpoint.
ln(1/2): Set x = 1/2 in ln(1 − x) = −∑xⁿ/n to get ln(1/2) = −1/2 − 1/8 − 1/24 − …
arctan(x): Integrate 1/(1 + x²) = ∑(−1)ⁿ x^(2n) to get:
tan⁻¹(x) = ∑(−1)ⁿ x^(2n+1)/(2n + 1), −1 ≤ x ≤ 1
At x = −1, the series for ln(1 − x) is ∑(−1)ⁿ/n, which converges (to ln 2). But the series for its derivative 1/(1 − x) at x = −1 is ∑(−1)ⁿ, which diverges. Differentiation can lose convergence at an endpoint; integration can gain it.
ln(1.1) = ln(1 − (−0.1)) = 0.1 − (0.01)/2 + (0.001)/3 − (0.0001)/4 + …
This is an alternating series, so ASET applies. The fifth term is 1/(5 · 10⁵) = 0.000002, so four terms give accuracy to five decimal places:
ln 1.1 ≈ 0.1 − 0.005 + 0.000333 − 0.000025 = 0.095308 ≈ 0.09531
eˣ = ∑(n=0 to ∞) xⁿ/n! = 1 + x + x²/2! + x³/3! + …
Converges for all x (R = ∞). Differentiating term by term returns the same series, confirming that (eˣ)′ = eˣ.
1/(1 − x) = ∑xⁿ, |x| < 1
1/(1 + x²) = ∑(−1)ⁿ x^(2n), |x| < 1
ln(1 − x) = −∑(n=1 to ∞) xⁿ/n, −1 ≤ x < 1
tan⁻¹(x) = ∑(n=0 to ∞) (−1)ⁿ x^(2n+1)/(2n + 1), −1 ≤ x ≤ 1
eˣ = ∑(n=0 to ∞) xⁿ/n!, all x
Term-by-term differentiation: f(x) = ∑cₙ(x−a)ⁿ ⟹ f′(x) = ∑n·cₙ(x−a)^(n−1), same R.
Term-by-term integration: ∫f(x)dx = C + ∑cₙ(x−a)^(n+1)/(n+1), same R.
⚠️ The ratio test finds R but tells you nothing about the endpoints. You must check each endpoint with a separate test (AST, p-series, divergence test, etc.).
⚠️ When deriving a series by substitution, the radius of convergence changes to match the substitution. If you replace x with −x/3, the condition |x| < 1 becomes |−x/3| < 1, i.e. |x| < 3.
⚠️ The convention is to expand around a = 0 (Maclaurin-style) unless told otherwise. If the problem says "around a = −2," your variable is (x + 2), not x.
⚠️ Integration can make a series converge at an endpoint where it previously diverged. Differentiation can do the opposite. The radius R stays the same, but the interval I may change.
⚠️ Questions like "for which x does this converge?" expect the full interval of convergence, endpoints included or excluded with justification.
⚠️ If you're told a power series converges at one point and diverges at another, you can deduce bounds on R and settle convergence for any x inside or outside those bounds. Points between the known convergent and divergent distances from the centre are inconclusive.
Q: What is the radius of convergence of ∑(2ⁿ/n²)xⁿ?
A: Ratio test gives L = 2|x|. Setting L < 1 gives |x| < 1/2, so R = 1/2.
Q: Find the interval of convergence of ∑(−1)ⁿ(x − 2)ⁿ/n.
A: Ratio test gives |x − 2| < 1, so R = 1. At x = 3: alternating harmonic, converges. At x = 1: harmonic series, diverges. I = (1, 3].
Q: Write the power series for 1/(1 + x²) centred at 0 and state its radius of convergence.
A: Substitute −x² for x in ∑xⁿ: 1/(1 + x²) = ∑(−1)ⁿ x^(2n), R = 1.
Q: Derive the power series for ln(1 − x) from the geometric series.
A: Integrate 1/(1 − x) = ∑xⁿ term by term: −ln(1 − x) = ∑xⁿ⁺¹/(n + 1). Set x = 0 to find C = 0. So ln(1 − x) = −∑xⁿ/n for n ≥ 1, valid for −1 ≤ x < 1.
Q: If ∑cₙ(x − 4)ⁿ converges at x = 2 and diverges at x = 8, what can you say about convergence at x = 5?
A: x = 2 is distance 2 from centre 4, so R ≥ 2. Since 5 is distance 1 from centre 4, and 1 < 2 ≤ R, the series converges at x = 5.
Q: Why does term-by-term differentiation of ∑xⁿ/n not work at x = −1?
A: ∑(−1)ⁿ/n converges (AST), but its derivative series ∑(−1)ⁿ diverges. Differentiation preserved R but lost convergence at the endpoint.
power series, radius of convergence, interval of convergence, geometric series, 1/(1−x), term-by-term differentiation, term-by-term integration, Maclaurin series, ln(1−x), arctan series, exponential series, ratio test for power series, endpoint convergence, MATH 308, calculus II, Texas A&M, infinite polynomial