Radius and Interval of Convergence

A series with x in it

A power series is a series whose terms are powers of x − a, each multiplied by a coefficient: $\sum c_n{\left(x-a\right)}^n=c_0+c_1\left(x-a\right)+c_2{\left(x-a\right)}^2+\dots$Σ cₙ(x − a)ⁿ = c₀ + c₁(x − a) + c₂(x − a)² + …. The number a is its center. Each value of x gives a different series of numbers, and some of those series converge while others do not.

Take $\sum \frac{x^n}{n}=x+\frac{x^2}{2}+\frac{x^3}{3}+\dots$Σ xⁿ/n = x + x²/2 + x³/3 + …. At $x=\frac{1}{2}$x = 1/2 the terms are 0.5, 0.125, 0.041667, 0.015625, 0.00625, …, they shrink quickly, and the series converges (its sum is ln 2 = 0.693147). At x = 2 the terms are 2, 2, 2.666667, 4, 6.4, …, they grow, and the series diverges.

At the center itself every term after the first is 0, so a power series always converges at x = a. The question is how far from a it keeps converging.

The ratio test, with x inside

Apply the ratio test to the terms $c_n{\left(x-a\right)}^n$cₙ(x − a)ⁿ. Dividing one term by the one before leaves one factor of x − a, so the size of the ratio is $|x-a|\times |c_{n+1}/c_n|$|x − a| × |cₙ₊₁ / cₙ|. The first factor does not depend on n. If $|c_{n+1}/c_n|$|cₙ₊₁ / cₙ| approaches a number k as n grows, then L = k|x − a|.

The series converges when L < 1, which is when $|x-a|<\frac{1}{k}$|x − a| < 1/k. That number is the radius of convergence, $R=\frac{1}{k}$R = 1/k. When |x − a| > R, L is above 1 and the series diverges.

An interval on the number line

|x − a| < R says that x is less than R away from a. On the number line that is every number from a − R to a + R, not counting the two ends: one stretch, centered on a, with a length of 2R.

A center at 2 and a radius of 3. The series converges for every x less than 3 away from 2, which is every x between −1 and 5.

The radius of $\sum \frac{x^n}{n}$Σ xⁿ/n

The terms are $\frac{x^n}{n}$xⁿ/n. Dividing the next term by this one gives $x\times \frac{n}{n+1}$x × n/(n + 1), whose size is $|x|\times \frac{n}{n+1}$|x| × n/(n + 1). As n grows, $\frac{n}{n+1}$n/(n + 1) tends to 1, so L = |x|.

L < 1 means |x| < 1. The center is 0 and the radius is R = 1, so the series converges for −1 < x < 1 and diverges for |x| > 1. That agrees with the two tries: $x=\frac{1}{2}$x = 1/2 is inside, and x = 2 is outside.

The two endpoints

At x = 1 and x = −1, L = |x| = 1, and the ratio test decides nothing. You can see it in the ratio itself: at an endpoint it is $\frac{n}{n+1}$n/(n + 1), which is 0.5 at n = 1, 0.909091 at n = 10 and 0.990099 at n = 100, creeping up to 1. So each endpoint needs its own test.

At x = 1 the series is $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots$1 + 1/2 + 1/3 + 1/4 + …, the harmonic series. It is a p-series with p = 1, so it diverges, and x = 1 is not in the interval.

At x = −1 the series is $-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\dots$−1 + 1/2 − 1/3 + 1/4 − …. The signs alternate, and the sizes 1, $\frac{1}{2}$1/2, $\frac{1}{3}$1/3, … decrease and tend to 0, so the alternating series test says it converges (its sum is −ln 2 = −0.693147). So x = −1 is in the interval.

The interval of convergence is $-1\le x<1$−1 ≤ x < 1: the left end is included and the right end is not. In bracket notation it is written [−1, 1), with a square bracket at an included end and a round one at an excluded end.

The interval of convergence of $\sum \frac{x^n}{n}$Σ xⁿ/n. The filled circle at −1 is included, because the alternating series there converges. The hollow circle at 1 is not, because the harmonic series diverges.

Four ways to end

The radius fixes where the interval stops, but not whether its ends belong to it. Four series with center 0 and radius 1 show every possibility.

$\sum x^n$Σ xⁿ is the geometric series. At x = 1 its terms are all 1, and at x = −1 they are 1, −1, 1, −1, …; neither tends to 0, so both ends diverge and the interval is −1 < x < 1.

$\sum \frac{x^n}{n^2}$Σ xⁿ/n² converges at x = 1, where it is the p-series with p = 2, and at x = −1, where the sizes are the same $\frac{1}{n^2}$1/n². Both ends are included: $-1\le x\le 1$−1 ≤ x ≤ 1.

$\sum \frac{x^n}{n}$Σ xⁿ/n gives $-1\le x<1$−1 ≤ x < 1, as above. Replace x by −x, and $\sum \frac{{\left(-x\right)}^n}{n}$Σ (−x)ⁿ/n swaps the two ends: it alternates at x = 1 and is the harmonic series at x = −1, so its interval is $-1<x\le 1$−1 < x ≤ 1.

A center away from 0

Take the series whose nth term is ${\left(x-2\right)}^n$(x − 2)ⁿ divided by $n^2\times 3^n$n² × 3ⁿ. Dividing the next term by this one gives $\frac{x-2}{3}\times \frac{n^2}{{\left(n+1\right)}^2}$(x − 2)/3 × n²/(n + 1)², and $\frac{n^2}{{\left(n+1\right)}^2}$n²/(n + 1)² tends to 1, so L is |x − 2| divided by 3.

L < 1 means |x − 2| < 3. The center is 2 and the radius is 3, so the series converges for −1 < x < 5.

Now the ends. At x = 5, x − 2 = 3, and the terms are $3^n$3ⁿ divided by $n^2\times 3^n$n² × 3ⁿ, which is $\frac{1}{n^2}$1/n²: a p-series with p = 2, which converges. At x = −1, x − 2 = −3, and the terms are $\frac{{\left(-1\right)}^n}{n^2}$(−1)ⁿ/n², whose sizes are again $\frac{1}{n^2}$1/n², so this converges too. Both ends are included, and the interval is $-1\le x\le 5$−1 ≤ x ≤ 5.

The interval of convergence for the series centered at 2 with radius 3. Both circles are filled, because at each end the sizes of the terms are $\frac{1}{n^2}$1/n², which add up to a finite total.

An infinite radius, and a radius of 0

For $\sum \frac{x^n}{n!}$Σ xⁿ/n!, the ratio is $\frac{x}{n+1}$x/(n + 1), whose size is |x| divided by n + 1. Whatever x is, this tends to 0 as n grows, so L = 0 for every x. The series converges for every real number, and the radius is said to be infinite. This is the series for $e^x$eˣ.

For $\sum n!x^n$Σ n! xⁿ, the ratio is (n + 1)x, whose size grows without limit for every x except 0. The series converges only at its center, and the radius is 0.

The usual mistakes

Stopping at the radius. R = 1 for $\sum \frac{x^n}{n}$Σ xⁿ/n does not make the interval −1 < x < 1 or $-1\le x\le 1$−1 ≤ x ≤ 1. Each end must be tested on its own, and here one is in and one is out.

Testing an endpoint with the ratio test. At an endpoint L = 1 exactly, which is the one value the ratio test cannot decide. Use the nth term test, a p-series, a comparison or the alternating series test there.

Forgetting the center. |x − 2| < 3 means −1 < x < 5, from 2 − 3 to 2 + 3, not −3 < x < 3.

A model with a limited range

In the application below, the energy in a stretched spring is given as a power series. The ratio test gives its radius, the two ends are tested with the nth term test, and the series is then summed at a point inside the interval and refused at a point outside it.