Radius and Interval of Convergence of Power Series: Study Guide (Calculus BC 10.13)

BC-only material. Power series are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

This topic uses almost every series test from earlier in Unit 10. If any is shaky, revisit it from the Calculus BC roadmap first. The previous topic, the Lagrange error bound, finished the work on Taylor polynomials.

Prerequisite	Topic	What you use it for here
Geometric series	10.2	The simplest power series; a model for everything else
nth term test	10.3	Showing divergence at an endpoint
Harmonic series and p-series	10.5	Deciding endpoints such as Σ 1/n or Σ 1/n²
Alternating series test	10.7	Deciding endpoints where the signs alternate
Ratio test	10.8	Finding the radius of convergence
Absolute and conditional convergence	10.9	Describing how the series converges at an endpoint
Taylor polynomials	10.11	Linking coefficients to derivatives

Notation on this page: Σ from n = 0 to ∞ of uₙ is the infinite series u₀ + u₁ + u₂ + … . We use c for the centre of a power series and R for the radius. The course description writes the centre as r; it means the same thing.

What a power series is

A power series centred at c is a series of the form

Σ from n = 0 to ∞ of aₙ(x − c)ⁿ = a₀ + a₁(x − c) + a₂(x − c)² + a₃(x − c)³ + …

Here n counts up from 0 through the whole numbers, the coefficients a₀, a₁, a₂, … are real numbers, and c is a fixed real number. By convention (x − c)⁰ = 1, even when x = c.

A power series looks like a polynomial that never ends. The difference is important: a polynomial gives a number for every x, but an infinite sum might not. For each value of x you substitute, you get an ordinary series of numbers, and that series either converges or diverges.

Example. For Σ from n = 0 to ∞ of xⁿ (centre 0, every coefficient 1):

• at x = ½ you get 1 + ½ + ¼ + …, a geometric series with sum 2;
• at x = 3 you get 1 + 3 + 9 + …, which diverges.

So the question for this topic is: for which x does the series converge?

One value is always safe. At x = c every term after the first is 0, so the series converges to a₀.

Only three things can happen

For any power series centred at c, exactly one of these is true:

1. It converges only at x = c. We say R = 0.
2. It converges for every real x. We say R = ∞.
3. There is a positive number R, the radius of convergence, such that the series converges (absolutely) when |x − c| < R and diverges when |x − c| > R.

In case 3 the open interval (c − R, c + R) is always inside the set where the series converges. The two endpoints, x = c − R and x = c + R, are not decided by R. Each one might converge or diverge, so there are four possible intervals of convergence: (c − R, c + R), [c − R, c + R), (c − R, c + R] and [c − R, c + R].

So the set of x where a power series converges is never scattered. It is a single point, a single interval centred at c, or the whole real line.

A useful consequence. Suppose a series centred at 1 converges at x = 4 and diverges at x = −5. Convergence at distance 3 from the centre means R ≥ 3. Divergence at distance 6 means R ≤ 6. So the series must converge at x = 3.5 (distance 2.5) and must diverge at x = 8 (distance 7), but x = −2 and x = 6 (distances 3 and 5) could go either way.

Finding the radius with the ratio test

The ratio test (Topic 10.8) works well because consecutive terms of a power series share almost everything. Let uₙ = aₙ(x − c)ⁿ be the whole term, including the x part.

1. Form |uₙ₊₁ / uₙ| and simplify. The powers of (x − c) cancel down to a single |x − c|.
2. Take the limit as n → ∞, keeping x fixed. Call it L. It will be |x − c| times a number, or 0, or ∞.
3. The series converges when L < 1 and diverges when L > 1. Solve L < 1 for |x − c|.
4. Read off R from |x − c| < R.

The three cases match the three outcomes above:

• L = (number) · |x − c|. Then L < 1 gives |x − c| < R for a positive R.
• L = 0 for every x. Then the series converges everywhere: R = ∞. Example: for Σ xⁿ/n!, |uₙ₊₁/uₙ| = |x|/(n + 1) → 0.
• L = ∞ for every x ≠ c. Then the series converges only at the centre: R = 0. Example: for Σ n! xⁿ, |uₙ₊₁/uₙ| = (n + 1)|x| → ∞ when x ≠ 0.

Only even powers? If the series is Σ x^(2n)/9ⁿ, the ratio is |x|²/9. Then |x|²/9 < 1 gives |x| < 3, so R = 3. Work with the whole term and you will not go wrong.

Worked example 1: a radius, then the endpoints

Question. Find the radius and the interval of convergence of Σ from n = 1 to ∞ of (x + 1)ⁿ / (n · 4ⁿ).

1. Identify the centre. x + 1 = x − (−1), so c = −1.
2. Ratio of consecutive terms. |uₙ₊₁ / uₙ| = |(x + 1)ⁿ⁺¹ / ((n + 1) · 4ⁿ⁺¹)| · |(n · 4ⁿ) / (x + 1)ⁿ| = (|x + 1| / 4) · n/(n + 1).
3. Limit. n/(n + 1) → 1, so L = |x + 1| / 4.
4. Solve L < 1. |x + 1| < 4, so R = 4. The series converges on −5 < x < 3 and diverges when |x + 1| > 4.
5. Right endpoint, x = 3. Then x + 1 = 4 and the series is Σ 4ⁿ/(n · 4ⁿ) = Σ 1/n, the harmonic series. It diverges.
6. Left endpoint, x = −5. Then x + 1 = −4 and the series is Σ (−4)ⁿ/(n · 4ⁿ) = Σ (−1)ⁿ/n. The terms alternate in sign, 1/n decreases and 1/n → 0, so by the alternating series test it converges. (Only conditionally: the absolute values give the harmonic series.)

Answer. Radius 4. Interval of convergence [−5, 3).

Checks.

• Centre and symmetry: −5 and 3 are both 4 units from −1. ✓
• A point inside: at x = 0 the series is Σ (1/4)ⁿ/n. Its terms are smaller than those of the convergent geometric series Σ (1/4)ⁿ, so it converges, as expected for a point inside.
• Notation: a square bracket at −5 (included), a round bracket at 3 (excluded).

Worked example 2: when x has a coefficient

Question. Find the radius and the interval of convergence of Σ from n = 1 to ∞ of (2x − 5)ⁿ / n².

1. Ratio. |uₙ₊₁ / uₙ| = |2x − 5| · n²/(n + 1)².
2. Limit. n²/(n + 1)² → 1, so L = |2x − 5|.
3. Solve L < 1. |2x − 5| < 1. This is not yet in the form |x − c| < R. Factor out the 2: |2x − 5| = 2|x − 5/2|, so 2|x − 5/2| < 1, which gives |x − 5/2| < 1/2.
4. Read off. Centre 5/2, R = 1/2. The open interval is (2, 3).
5. Right endpoint, x = 3. 2x − 5 = 1, so the series is Σ 1/n², a p-series with p = 2 > 1. It converges.
6. Left endpoint, x = 2. 2x − 5 = −1, so the series is Σ (−1)ⁿ/n². The series of absolute values is Σ 1/n², which converges, so this one converges absolutely.

Answer. Radius 1/2. Interval of convergence [2, 3].

Why the factoring matters. A common slip is to stop at |2x − 5| < 1 and say R = 1. But the interval (2, 3) has length 1, so the radius is half of that. Another way to see it: (2x − 5)ⁿ = 2ⁿ(x − 5/2)ⁿ, so the series is Σ [2ⁿ/n²](x − 5/2)ⁿ, a power series centred at 5/2.

Check. Both endpoints are 1/2 from 5/2. ✓

Endpoint toolkit

At an endpoint, x is a number, so you have an ordinary numerical series. Choose the test by its shape.

Endpoint series looks like	Test to use	Typical outcome
Σ 1/nᵖ	p-series	converges if p > 1, diverges if p ≤ 1
Σ (−1)ⁿ bₙ with bₙ decreasing to 0	alternating series test	converges
Terms that do not tend to 0	nth term test	diverges
Σ rⁿ	geometric series	converges only if |r| < 1
Close to a known series	comparison or limit comparison	same as the known series

The ratio test is missing from the table on purpose. At an endpoint its limit is exactly 1, which tells you nothing.

Power series and Taylor series

When a power series has a positive radius, its sum is a function on the open interval (c − R, c + R):

f(x) = a₀ + a₁(x − c) + a₂(x − c)² + … for |x − c| < R.

A key fact is that this power series is the Taylor series for f about c. In other words, the coefficients are tied to the derivatives:

aₙ = f⁽ⁿ⁾(c) / n!, so f⁽ⁿ⁾(c) = n! · aₙ

You built Taylor polynomials from derivatives in Topic 10.11. Here you can run the idea backwards: read a coefficient and get a derivative.

Example. Let h(x) = Σ from n = 0 to ∞ of xⁿ/(n + 2). The ratio test gives |x| · (n + 2)/(n + 3) → |x|, so R = 1 and h is defined for |x| < 1. The coefficient of x⁴ is a₄ = 1/6, so h⁽⁴⁾(0) = 4! · 1/6 = 4. You do not need a formula for h to find it.

Example. The geometric series 1/(1 − x) = Σ xⁿ for |x| < 1 has every coefficient equal to 1. Differentiating 1/(1 − x) repeatedly gives 1, 1, 2, 6, 24, … at x = 0, which is n!. Then n!/n! = 1 for every n, matching the coefficients. ✓

Topic 10.14 uses this link to build Taylor series directly.

Differentiating and integrating term by term

Inside its open interval, a power series can be differentiated or integrated one term at a time, like a polynomial. The new series has the same radius of convergence. But the endpoints can behave differently, so you must test them again.

Example. Let f(x) = Σ from n = 1 to ∞ of xⁿ/n².

Series	Radius	x = 1	x = −1	Interval
f(x) = Σ xⁿ/n²	1	Σ 1/n² converges	Σ (−1)ⁿ/n² converges	[−1, 1]
f′(x) = Σ xⁿ⁻¹/n	1	Σ 1/n diverges	Σ (−1)ⁿ⁻¹/n converges	[−1, 1)
f″(x) = Σ (n − 1)xⁿ⁻²/n	1	terms → 1, diverges	terms do not → 0, diverges	(−1, 1)

Each derivative multiplies the coefficient by n, which makes the endpoint series larger, so endpoints can be lost. Integrating divides by n + 1, so endpoints can be gained. The radius never changes.

Common misconceptions

• “The radius is the length of the interval.” The interval (c − R, c + R) has length 2R.
• Reading R straight from |2x − 5| < 1. Factor first: |x − 5/2| < 1/2 gives R = 1/2, not 1.
• Using the ratio test at the endpoints. The limit is 1 there, so the test is inconclusive. Use a different test.
• Forgetting to test the endpoints at all. The radius only gives the open interval. The course expects both endpoints decided, with a reason.
• Getting the centre’s sign wrong. (x + 1)ⁿ is centred at −1, not 1.
• Leaving x out of the ratio. If you take the limit of |aₙ₊₁/aₙ| on the coefficients alone, the result is 1/R, not R. Working with the whole term avoids this.
• “Conditional convergence means divergence.” Σ (−1)ⁿ/n converges, so x = −5 belongs in the interval in Worked example 1.
• Assuming the endpoints carry over after differentiating or integrating. Only the radius is guaranteed to stay the same.

Where this leads

The next topic, Finding Taylor or Maclaurin series for a function, builds full series for eˣ, sin x, cos x and 1/(1 − x), and you will state where each one converges using the methods on this page. Topic 10.15 then makes new series by substitution and term-by-term calculus, where the radius rule above saves time. Use the Calculus BC roadmap to see the order.

Try the practice questions now, then use the revision notes and the checklist to consolidate.