Power Series & Interval of Convergence

You’ll be able to

Find the radius of convergence using the ratio test
Determine the interval of convergence, checking the endpoints
Understand term-by-term differentiation and integration of power series

A power series is a polynomial of infinite degree

A power series centered at a has the form Σcₙ(x − a)ⁿ. Whether it converges depends on x: it always converges at the center x = a, and typically on a symmetric interval around it. The half-width of that interval is the radius of convergence R. Inside |x − a| < R the series converges absolutely; outside it diverges; and exactly at the two endpoints the behavior must be checked separately.

Radius of convergence via the ratio test

lim |cₙ₊₁(x − a)ⁿ⁺¹ / cₙ(x − a)ⁿ| < 1 ⟹ |x − a| < R

Apply the ratio test, solve the resulting inequality for |x − a|; the bound is R. Then test x = a ± R by hand.

The interval of convergence and its endpoints

The interval of convergence is the full set of x for which the series converges. The ratio test gives the open interval (a − R, a + R), but it is silent at the endpoints (there the ratio limit equals 1). So you must substitute x = a − R and x = a + R individually and apply another test — often the alternating series test or the p-series test. Each endpoint may independently converge or diverge, so the interval can be open, closed, or half-open.

Differentiating and integrating term by term

Within its interval of convergence, a power series can be differentiated and integrated term by term, and the new series has the same radius R (though endpoint behavior may change). This is powerful: integrating the geometric series 1/(1 − x) = Σxⁿ term by term yields the series for −ln(1 − x); differentiating shifts powers down. It is how many standard series are derived from the handful you memorize.

Worked example

Find the radius of convergence of Σ (n=1 to ∞) xⁿ / n.

1.Apply the ratio test: |aₙ₊₁/aₙ| = |xⁿ⁺¹/(n+1)| · |n/xⁿ| = |x|·n/(n+1).
2.Take the limit as n → ∞: n/(n+1) → 1, so the limit is |x|.
3.Convergence requires |x| < 1, so the radius of convergence is R = 1.
4.(Endpoints: at x = 1 it is the divergent harmonic series; at x = −1 it is the convergent alternating harmonic series, giving interval [−1, 1).)

Answer: R = 1, with interval of convergence [−1, 1) once the endpoints are checked — convergent at −1, divergent at 1.

Watch out

The ratio test never decides the endpoints — there the limit is exactly 1, the inconclusive case. You must plug each endpoint in separately and run a different test. Reporting an open interval without checking endpoints is an incomplete answer.

Checkpoint

A power series Σcₙ(x − 2)ⁿ has radius of convergence R = 3. On which open interval is convergence guaranteed?

Checkpoint

When you differentiate a power series term by term within its interval of convergence, the radius of convergence R:

On the exam

A complete interval-of-convergence answer has three parts: apply the ratio test for R, then test each endpoint separately, then state the interval with the correct brackets. Free-response rubrics award a point specifically for the endpoint analysis.

Answer the 2 checkpoints as you read.