Find the center and radius of convergence

Where a power series converges

The series \(1+\frac{x-2}{3}+\left(\frac{x-2}{3}\right)^2+\cdots\) is a geometric series with ratio \(\frac{x-2}{3}\), which changes with \(x\). The powers are powers of \(x-2\), so the center is 2.

Drag the orange point along the number line. The dots show the partial sums, which are the sums of the first terms. If the ratio is between −1 and 1, the dots level off at one value, such as 1.5 at \(x=3\).

Outside that range, the dots grow without limit or swing wider and wider, so the series diverges. Near an end of the range, the dots need many terms to level off.

Show the teal band. It covers every x less than 3 from the center, where the dots level off, so 3 is the radius of convergence.

The center and the radius

The center \(a\) is the input where \(x-a\) is 0. The teal band reaches the same distance \(R\) to each side of \(a\).

Move the point \(x\). Inside the band, the orange distance \(|x-a|\) is less than \(R\), and the series converges. Outside the band, the distance is more than \(R\), and the series diverges.

The size of each term depends on \(x\) only through \(|x-a|\), so the band is symmetric around \(a\).

Notation

1. A power series with center \(a\) is a sum of terms \(c_n(x-a)^n\). The numbers \(c_n\) are the coefficients.

   \[\sum_{n=0}^{\infty} c_n\,(x-a)^n\]

2. Divide the size of a term by the size of the term before it. The powers cancel except for one factor \(\cy{|x-a|}\).

   \[\begin{aligned}&\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| \\ &\quad=\left|\frac{c_{n+1}}{c_n}\right|\,\cy{|x-a|}\end{aligned}\]

3. Let \(K\) be the limit of the coefficient ratio \(\left|\frac{c_{n+1}}{c_n}\right|\) as \(n\) grows. By the ratio test, the series converges if \(K|x-a|\lt 1\), because the terms then shrink like those of a geometric series with ratio less than 1.

   \[K\,\cy{|x-a|}\lt 1\]

   If \(K|x-a|>1\), the terms grow and the series diverges.

4. If \(K\) is positive, divide both sides by \(K\). The distance must be less than \(\cb{R}=\frac1K\), the radius of the band.

   \[\cy{|x-a|}\lt \cb{R},\qquad \cb{R}=\frac{1}{K}\]

5. If \(K\) is 0, the product \(K|x-a|\) is 0 for every \(x\), so \(R\) is infinite and the series converges for all \(x\).

   \[K=0\ \Rightarrow\ \cb{R}=\infty\]

6. If the coefficient ratio grows without bound, the product exceeds 1 for every \(x\) except \(a\). Then \(R\) is 0, and the series converges only at the center.

   \[K=\infty\ \Rightarrow\ \cb{R}=0\]

7. The ratio test cannot decide the two ends of the band, where the product equals 1. The next skill tests the ends.

When it fails

The ratio test needs every coefficient to be nonzero. The series \(1+\frac{x^2}{4}+\frac{x^4}{16}+\cdots\) has only even powers, so written with every power it has the terms \(0\cdot x\) and \(0\cdot x^3\), and you cannot divide by them.

Instead, divide each nonzero term by the nonzero term before it. This ratio is \(\frac{x^2}{4}\), which is less than 1 when \(|x|\lt 2\), so the radius is 2.

Example 1: A factor in n and a factor in x

Find the center and the radius of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{n\,(x+1)^n}{5^n}\).

1. The powers are powers of \(x+1\), which is \(x-(-1)\), so the center is −1.

   \[(x+1)^n=\bigl(x-(-1)\bigr)^n\]

2. Divide the size of a term by the size of the term before it.

   \[\begin{aligned}&\frac{(n+1)\,|x+1|^{n+1}/5^{n+1}}{n\,|x+1|^{n}/5^{n}} \\ &\quad=\frac{n+1}{n}\cdot\frac{\cy{|x+1|}}{5}\end{aligned}\]

3. As \(n\) grows, \(\frac{n+1}{n}\) approaches 1, so the limit of the ratio is \(\frac{|x+1|}{5}\).

   \[\frac{n+1}{n}\to 1\]

4. The series converges when this limit is less than 1. Multiply both sides by 5.

   \[\cy{|x+1|}\lt \cb{5}\]

5. The radius is 5, so the series converges for \(x\) between −6 and 4, the two points 5 from the center.

   \[-6\lt x\lt 4\]

Example 2: A limit of 0

Find the center and the radius of convergence of \(\displaystyle\sum_{n=0}^{\infty}\frac{(x-3)^n}{n!}\).

1. The center is 3. Because \((n+1)!=(n+1)\,n!\), the factorials leave one factor \(n+1\) in the ratio of consecutive terms.

   \[\frac{|x-3|^{n+1}/(n+1)!}{|x-3|^{n}/n!}=\frac{\cy{|x-3|}}{n+1}\]

2. A tempting wrong move: As \(n\) grows, the ratio approaches 0. The limit is 0, so the radius is 0.

   \[K=0\ \Rightarrow\ R=0\]

   At \(x=9\), 6 from the center, the dots level off at about 403.4, so the series converges there. A radius of 0 would allow only \(x=3\).

3. The limit 0 is less than 1 for every \(x\), so the radius is infinite.

   \[0\lt 1\ \text{for every }x\ \Rightarrow\ \cb{R}=\infty\]

Practice

Find the center and the radius of convergence of \(\displaystyle\sum_{n=0}^{\infty} n!\,(x-4)^n\).

In the ratio of consecutive terms, the factorials leave the factor \(n+1\). Drag the point away from 4 and see what the dots do as \(n\) grows.

Questions

Questions 1 and 2 use the figure. The teal band shows where a power series converges. Its ends are not tested.

1. What is the center of the series?

   • 1
   • 4
   • 5
   • 0
   Answer

   1

2. What does the figure show about the series at x = 6?

   • It diverges, because 6 is more than the radius from the center.
   • It converges, because 6 is close to the right end.
   • The ratio test cannot decide, because 6 is an end of the band.
   Answer

   It diverges, because 6 is more than the radius from the center.

3. What are the center and the radius of convergence of \(\displaystyle\sum_{n=0}^{\infty}\frac{5^n(x+1)^n}{n+2}\)?

   • Center −1, radius \(\frac15\)
   • Center −1, radius 5
   • Center 1, radius \(\frac15\)
   Answer

   Center −1, radius \(\frac15\)

Summary

You read the center a from the powers of x − a, and you find the radius R = 1/K from the limit K of the coefficient ratio.

This works because, closer to the center than R, the terms shrink like those of a geometric series with ratio less than 1.

You use it to find where a power series converges, before you test its two ends.

This is one of four skills in topic 10.13 of Which & Why: AP Calculus.