Power Series and the Interval of Convergence

A power series \(\sum c_n (x-a)^n\) is a series with a variable in it — so "does it converge?" becomes "for which x does it converge?" The answer is always an interval centered at \(a\): the radius of convergence R comes from the ratio test, and the two endpoints get checked by hand.

By the end you'll be able to run the ratio test on a power series to find its radius of convergence R, then test both endpoints \(a-R\) and \(a+R\) separately to write down the full interval of convergence.

Predict: grow R — does the convergence interval widen symmetrically about a, or stretch to one side? Slide R to check.

The shaded band is where the ratio test guarantees \(|x-a| < R\): convergence. The two hollow endpoint circles at \(a \pm R\) are flagged "check separately" — the ratio test gives exactly L = 1 there and says nothing. Drag the test point x across the line and watch the verdict flip; hover or tab to the center, endpoints, and test point for exact readouts.

|x − a| = 1.0 < R = 2.0 — the ratio test gives L < 1: converges absolutely at x = 4.0.

The number line — convergence interval \((a-R,\ a+R)\), its center, and the two endpoints the ratio test can't decide
converges (|x−a| < R) center a endpoints — check separately

A power series \(\sum c_n(x-a)^n\) converges on an interval centered at \(a\); the ratio test yields the radius of convergence R, and the endpoints must be checked separately to get the interval of convergence.

Visual

Picture the number line you just played with. A power series \(\sum c_n(x-a)^n\) carves it into three zones: a central band of width 2R around the center a where the series converges, everything outside where it diverges, and the two boundary points \(a-R\) and \(a+R\) where anything can happen. The band is always symmetric about \(a\) — because the ratio test's condition is \(|x-a| < R\), a distance condition — and the only suspense left is whether each endpoint is in or out. Every power series lands in one of three cases: the band shrinks to the single point \(x=a\) (R = 0), covers the whole line (R = ∞), or has some finite width (0 < R < ∞) — try sliding R to 0 in the demo to see the first case.

Formal

A power series centered at a is \(\sum_{n=0}^{\infty} c_n(x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots\), a function of x: plugging in a number for x produces an ordinary numerical series. To find where it converges, apply the ratio test — the one from the previous lesson — to the terms \(a_n = c_n(x-a)^n\): $$L = \lim_{n\to\infty}\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| = |x-a|\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|.$$ The series converges absolutely where \(L < 1\) and diverges where \(L > 1\); solving \(L < 1\) for x yields \(|x-a| < R\), which hands you R directly. At \(x = a \pm R\) the limit is exactly L = 1 — the ratio test's blind spot — so each endpoint's numerical series needs a direct test (alternating series test, p-series, comparison, …) before you can write the interval of convergence with the right brackets.

Worked example

Find the radius and interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-3)^n}{n\,2^n}\).

  1. Ratio test on \(a_n = \dfrac{(x-3)^n}{n\,2^n}\), centered at \(a = 3\): $$L=\lim_{n\to\infty}\left|\frac{(x-3)^{n+1}}{(n+1)2^{n+1}}\cdot\frac{n\,2^n}{(x-3)^n}\right| =\lim_{n\to\infty}\frac{n}{n+1}\cdot\frac{|x-3|}{2}=\frac{|x-3|}{2}.$$
  2. Require \(L < 1\): \(\dfrac{|x-3|}{2} < 1 \iff |x-3| < 2\). So R = 2, and the series converges absolutely for \(1 < x < 5\) — exactly the picture above with a = 3, R = 2.
  3. Left endpoint \(x = 1\): the series becomes \(\sum \dfrac{(-2)^n}{n\,2^n} = \sum\dfrac{(-1)^n}{n}\), the alternating harmonic series — converges. Include x = 1.
  4. Right endpoint \(x = 5\): the series becomes \(\sum \dfrac{2^n}{n\,2^n} = \sum\dfrac{1}{n}\), the harmonic series — diverges. Exclude x = 5.
  5. Interval of convergence: \([1, 5)\).
Your turn

Same procedure, new numbers: \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-1)^n}{n\,4^n}\). The ratio test gives \(L = \dfrac{|x-1|}{4}\), so \(L < 1\) means \(|x-1| < 4\): R = 4 and the open interval is \((-3, 5)\). Now finish it: at \(x = -3\) the series becomes \(\sum\frac{(-4)^n}{n\,4^n} = \) ____, which ____; at \(x = 5\) it becomes \(\sum\frac{4^n}{n\,4^n} = \) ____, which ____. Interval of convergence: ____

Reveal the answer

At \(x=-3\): \(\sum\frac{(-4)^n}{n\,4^n} = \sum\frac{(-1)^n}{n}\), the alternating harmonic series — converges, so include −3. At \(x=5\): \(\sum\frac{4^n}{n\,4^n} = \sum\frac{1}{n}\), the harmonic series — diverges, so exclude 5. Interval of convergence: \([-3, 5)\). Set a = 1 and R = 4 in the demo above and check the endpoint flags sit at −3 and 5.

More info — the three things a power series can do

Every power series does exactly one of three things: (1) converges only at its center \(x = a\) — that's R = 0, as with \(\sum n!\,x^n\), whose factorial coefficients grow too fast for any other x; (2) converges for every real x — R = ∞, no endpoints to check; or (3) converges when \(|x-a| < R\) and diverges when \(|x-a| > R\) for some finite R > 0 — the only case with endpoint suspense. The simplest finite-R example is the geometric series \(\sum x^n\) (center 0, all coefficients 1), which converges exactly when \(|x| < 1\). The OpenStax section linked in Dive deeper below works one ratio-test example for each of the three cases.

Check your understanding

Question 1 of 4

Find the radius of convergence of \(\displaystyle\sum_{n\ge 1} \frac{(x+2)^n}{n\,5^n}\).

Question 2 of 4

The ratio test on a power series gives \(|x-a| < R\). Why must the endpoints \(x = a-R\) and \(x = a+R\) still be checked separately?

Question 3 of 4

Apply the ratio test to \(\displaystyle\sum_{n\ge 0} n!\,x^n\). What do you conclude?

Question 4 of 4

A power series centered at \(a = 4\) has radius of convergence \(R = 3\). Direct endpoint tests show it diverges at \(x = 1\) and converges at \(x = 7\). What is its interval of convergence?

Recap

  • A power series \(\sum c_n(x-a)^n\) is a function of x; it converges on an interval centered at its center \(a\).
  • The ratio test on \(c_n(x-a)^n\) gives \(L = |x-a|\lim|c_{n+1}/c_n|\); solving \(L < 1\) yields \(|x-a| < R\) — the radius of convergence.
  • Three cases: R = 0 (converges only at x = a), R = ∞ (converges everywhere), or a finite R with convergence on \((a-R,\ a+R)\).
  • At the endpoints \(x = a \pm R\) the ratio test gives L = 1 and is inconclusive — test each endpoint's numerical series directly to decide the brackets on the interval of convergence.

Dive deeper

Sources

  • Power Series, Radius of Convergence, and Interval of Convergence