Test each finite endpoint separately
The two ends of the band
The series \(\sum_{n=1}^{\infty}\frac{(x-1)^n}{n\,2^n}\) has center 1 and radius 2. By the ratio test, it converges for every \(x\) in the teal band from −1 to 3 and diverges outside it. The open rings mark the two ends, where the test gives no answer. The teal dots below show the partial sums of the series at the orange point \(x\).
Press the button for the right end. At \(x=3\) every term is \(\frac1n\), and the dots rise slowly without limit, so the series diverges.
Press the button for the left end. At \(x=-1\) the signs alternate, and the dots move toward the dashed line at −0.693 by smaller steps, so the series converges.
The two ends of one band behave in different ways, so each end needs its own test.
Each end is its own series
The band has two ends, at \(a-R\) and \(a+R\), and both are open rings because they are not tested yet.
If you replace x by an end, the power series becomes a series of numbers. Press a button to test one end. A filled ring means that the series converges at that end, and an open ring means that it diverges.
Each end has its own series and its own result.
Notation
Find the two ends from the center \(a\) and the radius \(R\).
\[x=\cy{a-R},\qquad x=\cy{a+R}\]Replace x by one end in the original series. The result is a series of numbers, which the tests below can decide.
If the terms do not approach 0, the series diverges.
\[\lim_{n\to\infty}a_n\neq 0\ \Rightarrow\ \text{diverges}\]If the terms are \(\frac{1}{n^p}\), the series is a p-series. It converges when \(p>1\) and diverges when \(p\le 1\).
\[\sum\frac{1}{n^{p}}\ \begin{cases}\text{converges} & p>1 \\ \text{diverges} & p\le 1\end{cases}\]If the signs alternate and the sizes decrease to 0, the series converges by the alternating series test.
\[\begin{aligned}&\sum(-1)^n b_n,\quad b_n\downarrow 0 \\ &\quad\Rightarrow\ \text{converges}\end{aligned}\]Test the two ends one at a time, and write the result for each end.
When it fails
In the opening series, the ratio of the sizes of consecutive terms is \(\frac{n}{n+1}\) at both \(x=3\) and \(x=-1\).
Both ratios approach 1, but one series diverges and the other converges. A limit of 1 means that the ratio test gives no answer.
Example 1: Both ends
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{(x+2)^n}{n^2\,3^n}\) has center −2 and radius 3. Test both ends.
The ends are \(-2-3\) and \(-2+3\).
\[x=-5,\qquad x=1\]At \(x=1\), the terms are \(\frac{1}{n^2}\). This is a p-series with \(p=2\), so it converges.
\[\sum\frac{3^n}{n^2\,3^n}=\sum\frac{1}{n^2}\]At \(x=-5\), the terms are \(\frac{(-1)^n}{n^2}\). The signs alternate and the sizes \(\frac{1}{n^2}\) decrease to 0, so the series converges by the alternating series test.
\[\sum\frac{(-3)^n}{n^2\,3^n}=\sum\frac{(-1)^n}{n^2}\]The series converges at both ends.
Example 2: A tempting test
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{n\,(x-4)^n}{2^n}\) has center 4 and radius 2. Test both ends.
The ends are 2 and 6. At \(x=6\), the terms are \(n\).
\[\sum\frac{n\,2^n}{2^n}=\sum n\]The terms \(n\) do not approach 0, so the series diverges at \(x=6\).
A tempting wrong move: At \(x=2\), the terms are \((-1)^n n\), and their signs alternate. By the alternating series test, the series converges.
\[\sum(-1)^n\,n\ \text{converges}\]The alternating series test needs the sizes to decrease to 0, but the sizes \(n\) grow. The partial sums \(-1, 1, -2, 2, -3\) move farther from 0 each time.
The terms \((-1)^n n\) do not approach 0, because their sizes \(n\) grow, so the series diverges at \(x=2\). It diverges at both ends.
Practice
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{(x+1)^n}{\sqrt{n}}\) has center −1 and radius 1. Test both ends.
Drag the point to each end and compare the dots. Then choose a test that fits the signs of the terms.
Solution
The ends are −2 and 0.
\[x=-2,\qquad x=0\]At \(x=0\), the terms are \(\frac{1}{\sqrt{n}}=\frac{1}{n^{1/2}}\). This is a p-series with \(p=\frac12\), which is at most 1, so it diverges.
\[\sum\frac{1}{n^{1/2}}\]At \(x=-2\), the terms are \(\frac{(-1)^n}{\sqrt{n}}\). The signs alternate and the sizes decrease to 0, so the series converges.
\[\sum\frac{(-1)^n}{\sqrt{n}}\]The series converges at \(x=-2\) and diverges at \(x=0\).
Questions
The bars show the terms of a series at its two ends. The height of each bar is the size of its term. Green bars are positive terms and point up, and red bars are negative terms and point down.
What does the top row show about the series at the left end?
- It converges, because the signs alternate and the sizes decrease to 0.
- It diverges, because the bars change sign.
- It diverges, because there are infinitely many nonzero terms.
Answer
It converges, because the signs alternate and the sizes decrease to 0.
Can the same test show that the series converges at the right end?
- No. The bars do not alternate, and the terms 1/n form the harmonic series, which diverges.
- Yes, because the bars also shrink to 0.
- Yes, because the left end converges and the two ends match.
Answer
No. The bars do not alternate, and the terms 1/n form the harmonic series, which diverges.
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)}\) has center 0 and radius 1. At which ends does it converge?
- At both x = −1 and x = 1
- Only at x = 1
- Only at x = −1
Answer
At both x = −1 and x = 1
Summary
At each end, a − R and a + R, the power series becomes a series of numbers that needs its own test.
The ratio test gives a limit of 1 at both ends, so it cannot decide them, and the two ends can give different results.
You test the ends after you find the radius, to learn which ends belong to the interval where the series converges.
This is one of four skills in topic 10.13 of Which & Why: AP Calculus.