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The radius gives an open interval; each endpoint is then a separate series of numbers with its own test. The ratio test is necessarily silent at $|x - a| = R$, and the two ends need not agree.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can turn a radius of convergence into a full interval: writing down the two endpoints, substituting each into the series in turn, settling the numerical series that results with whichever test suits it, and assembling an interval whose two brackets were each decided separately. You can also say why the ratio test must be silent at an endpoint rather than merely happening to be.
You can find a radius of convergence: ratio test with the $x$ carried through, then solve the condition for the distance from the centre. That gives an open interval. This lesson is about its two ends, which the ratio test cannot reach and which are where all the interesting behaviour lives.
Endpoint: either of the two points exactly one radius from the centre.
Interval of convergence: every $x$ at which the series converges — the open interval, together with whichever endpoints pass their own test.
Closed at an end: that endpoint is included, written with a square bracket.
Open at an end: it is excluded, written with a round bracket.
The radius $R$ tells you that the series converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$. It says nothing about the two points where $|x - a| = R$, and that is not a weakness of the method.
At those two points the ratio test's limit is exactly $1$, which is the value at which the test is silent. It has to be silent there, because
$$\sum x^n, \qquad \sum \frac{x^n}{n}, \qquad \sum \frac{x^n}{n^2}$$
all have radius $1$ about $0$, and their intervals of convergence are $(-1, 1)$, $[-1, 1)$ and $[-1, 1]$. A test that spoke at the boundary would have to give one answer to three series that do three different things.
So an interval-of-convergence problem is three problems. One ratio test for the radius, then one ordinary series test at each endpoint — substitute the value, look at the series of numbers you are left with, and settle it with whatever suits: $p$-series, alternating, or the $n$th-term test.
The two ends need not agree. All four combinations of included and excluded occur, and the asymmetric ones are the common case. The reason is easy to see once stated: at one end the signs alternate and at the other they do not, and alternating is exactly the thing that can rescue a series whose terms shrink too slowly.
Answer the question that was asked. "Find the interval of convergence" is answered with an interval, brackets and all. A number is an answer to a different question.
Another way: picture
Picture the shaded segment from the last lesson, with a small circle at each end. The shading was settled in one calculation; each circle is filled in or left hollow by its own separate calculation afterwards. Two circles, two calculations, and no reason whatever for them to come out the same.
Another way: steps
Substituting an endpoint into $\sum c_n(x - a)^n$ always cancels the $(x-a)^n$ against whatever geometric factor the coefficients carry, and what is left is one of a very small number of familiar series.
| The endpoint series | What settles it | Verdict |
|---|---|---|
| $\sum 1$ or $\sum(-1)^n$ | $n$th-term test | diverges |
| $\sum \dfrac1n$ | $p$-series, $p = 1$ | diverges |
| $\sum \dfrac{(-1)^n}{n}$ | alternating series test | converges |
| $\sum \dfrac{1}{n^2}$ | $p$-series, $p = 2$ | converges |
| $\sum \dfrac{1}{\sqrt n}$ | $p$-series, $p = \tfrac12$ | diverges |
Being able to recognise those five on sight is most of what makes endpoint testing quick. The right-hand end usually gives the unsigned version and the left-hand end the alternating one, which is why the left end is included more often than the right.
Inferring the endpoints from the radius. The radius says nothing about them — that is what being an endpoint means here.
Assuming the two ends agree. They frequently do not, and the commonest series in the whole topic, $\sum \frac{x^n}{n}$, is one where they do not.
Stopping at the radius. The radius is produced by a routine calculation and so it feels like the result. The endpoints are fiddly and get dropped, which loses the marks and, more importantly, loses the part of the answer that distinguishes one series from another.
$\sum_{n=1}^{\infty}\dfrac{(x + 1)^{n}}{n}$: centre $-1$, and the ratio is $\dfrac{n}{n+1}|x + 1| \to |x + 1|$, so $R = 1$ and the open interval is $(-2, 0)$.
Radius first.
At $x = 0$: the series is $\sum \dfrac1n$, the harmonic series, which diverges. Excluded.
One endpoint, one test.
At $x = -2$: the series is $\sum \dfrac{(-1)^{n}}{n}$, which converges by the alternating series test. Included. The interval is $[-2, 0)$.
$\sum_{n=1}^{\infty}\dfrac{x^{n}}{n^{2}}$: the ratio is $\left(\dfrac{n}{n+1}\right)^{2}|x| \to |x|$, so $R = 1$ about $0$.
The square changes nothing.
At $x = 1$: $\sum \dfrac{1}{n^{2}}$, a $p$-series with $p = 2$. Converges, so included.
At $x = -1$: $\sum \dfrac{(-1)^{n}}{n^{2}}$, which converges absolutely by the same $p$-series. Included. The interval is $[-1, 1]$ — and that agreement was a conclusion, reached twice.
The ratio is $\dfrac{n}{n+1}\cdot\dfrac{|x-4|}{3} \to \dfrac{|x-4|}{3}$, so $|x - 4| < 3$: radius $3$ about $4$, open interval from $1$ to $7$.
Radius and centre.
At $x = 1$ the term is $\dfrac{(-3)^{n}}{n\,3^{n}} = \dfrac{(-1)^{n}}{n}$: alternating harmonic, which converges. Included.
Left end.
At $x = 7$ the term is $\dfrac{3^{n}}{n\,3^{n}} = \dfrac1n$: harmonic, which diverges. Excluded. So the interval is $[1, 7)$.
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x - 6)^{n}}{n\,2^{\,n}}$.
This task has no paper form; do it on a device.
Put the steps of finding the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x - 3)^{n}}{n^{2}}$ into the order you do them.
Number the steps in order (write the number in the box):
Each of these has radius $1$ about $0$. Say whether each endpoint is in the interval of convergence.
| At $x = -1$ | At $x = 1$ | |
|---|---|---|
| $\sum x^{n}$ | ||
| $\sum \dfrac{x^{n}}{n}$ | ||
| $\sum \dfrac{x^{n}}{n^{2}}$ |
A power series has radius $2$ about the centre $5$, and it is known to converge at $x = 3$. What follows about $x = 7$?
$\sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ has radius $1$ about the centre $0$. At which value of $x$ is its left-hand endpoint?
Answer:
A power series has interval of convergence from $-1$ to $3$, closed at the left end and open at the right. Give its centre, then its radius.
centre = m, radius = w
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x - 1)^{n}}{n\,4^{\,n}}$.
This task has no paper form; do it on a device.
You can produce a full interval of convergence, endpoints included. Without looking: how many separate tests does one interval-of-convergence problem need, and what do the three series of radius one about zero show?
9. Your turn: the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x - 4)^{n}}{n\,3^{\,n}}$, step 3