State the complete convergence interval

Marking the ends

The series \(\sum_{n=1}^{\infty}\frac{(-1)^n(x+3)^n}{n\,2^n}\) has center −3 and radius 2, so it converges for every x strictly between −5 and −1. This open range is the teal band.

At \(x=-5\) the terms are \(\frac1n\), the harmonic series, so the series diverges. At \(x=-1\) the terms are \(\frac{(-1)^n}{n}\), so the alternating series test shows that it converges.

Press the buttons to test each end. An open ring marks an end that is out of the interval, and a filled ring marks an end that is in it.

The ring at −5 stays open and the ring at −1 fills, so the complete interval is (−5, −1]. The round bracket leaves out −5, and the square bracket includes −1.

Four ways to finish the band

The teal band runs from \(a-R\) to \(a+R\), and every x inside it is in the interval. Each end is either in or out, so there are four possible intervals.

Press the four buttons. An open ring gives a round bracket, and a filled ring gives a square bracket.

In all four cases the band and its center \(a\) stay the same, and only the brackets change.

Notation

1. The center \(a\) and the radius \(R\) give the open range, where the series converges.

   \[a-R\lt x\lt a+R\]

2. Test each end on its own. The teal dots below show the partial sums at the orange point, which grow at −5 and approach a limit at −1. Write a round bracket at an end that is out and a square bracket at an end that is in.

   \[\begin{aligned}&(a-R,\ a+R),\ [a-R,\ a+R), \\ &(a-R,\ a+R],\ [a-R,\ a+R]\end{aligned}\]

3. If \(R\) is infinite, the series converges for every \(x\). If \(R\) is 0, the series converges only at the center.

   \[(-\infty,\ \infty),\qquad \{a\}\]

4. The center is halfway between the ends, so you can use it to check an interval.

   \[a=\frac{(a-R)+(a+R)}{2}\]

When it fails

Suppose you write \((-5,-1)\) for the opening series. The red \(\times\) marks −1, which this interval leaves out, but the partial sums at −1 approach a limit of about −0.693.

The series converges at −1, so the complete interval must include it: \((-5,-1]\).

Example 1: Both ends out

Write the complete interval of convergence of \(\displaystyle\sum_{n=0}^{\infty}\frac{(x-1)^n}{4^n}\).

1. This is a geometric series with ratio \(\frac{x-1}{4}\), so it converges when \(|x-1|\lt 4\). The center is 1 and the radius is 4, so the open range is from −3 to 5.

   \[-3\lt x\lt 5\]

2. At \(x=5\) every term is 1, and at \(x=-3\) the terms are 1 and −1 in turn. The terms do not approach 0 at either end, so the series diverges at both ends.

   \[\sum 1^n,\qquad \sum(-1)^n\]

3. Both ends are out, so both brackets are round.

   \[(-3,\ 5)\]

Example 2: Both ends in

Write the complete interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x+5)^n}{n^3\,2^n}\).

1. The ratio of the sizes of consecutive terms approaches \(\frac{|x+5|}{2}\). This is less than 1 when \(|x+5|\lt 2\), so the center is −5 and the radius is 2.

   \[\frac{n^3}{(n+1)^3}\cdot\frac{|x+5|}{2}\ \to\ \frac{|x+5|}{2}\]

2. A tempting wrong move: The radius is 2, so the interval is from −2 to 2.

   \[[-2,\ 2]\]

   The ends lie 2 from the center −5, not 2 from 0. The orange point at 0 is in \([-2,2]\), but the ratio there is \(\frac52\), which is more than 1, so the series diverges at 0.

3. The ends lie 2 on each side of −5, so the open range is from −7 to −3.

   \[-7\lt x\lt -3\]

4. At \(x=-3\) the terms are \(\frac{1}{n^3}\), a p-series with \(p=3\), so the series converges. At \(x=-7\) the terms are \(\frac{(-1)^n}{n^3}\), whose sizes form the same p-series, so the series converges absolutely. Both ends are in.

   \[[-7,\ -3]\]

Practice

Write the complete interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-7)^n}{n\,5^n}\).

Find the ends \(7-5\) and \(7+5\). At each end, ask whether the signs alternate.

Questions

The line shows where a power series converges. A filled ring marks an end where the series converges, and an open ring marks an end where it diverges.

1. How is the complete interval of convergence written?

   • \((-4,\,2]\)
   • \([-4,\,2)\)
   • \([-4,\,2]\)
   • \((-4,\,2)\)
   Answer

   \((-4,\,2]\)

2. Which of these inputs is not in the complete interval?

   • −4
   • 2
   • −1
   • 0
   Answer

   −4

3. What is the complete interval of convergence of \(\displaystyle\sum_{n=0}^{\infty}\frac{2^n(x-3)^n}{n+1}\)?

   • \(\left[\frac52,\,\frac72\right)\)
   • \(\left(\frac52,\,\frac72\right]\)
   • \(\left[\frac52,\,\frac72\right]\)
   • \(\left(\frac52,\,\frac72\right)\)
   Answer

   \(\left[\frac52,\,\frac72\right)\)

Summary

The interval of convergence contains the open range from \(a-R\) to \(a+R\), so only its two ends are in question.

Each end gives its own series, so each end takes a square bracket if its series converges and a round bracket if it diverges.

Write the interval this way whenever a question asks for the interval of convergence of a power series.

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