Back to the on-screen lesson ·
A power series converges on an interval about its centre; the ratio test finds the radius and is necessarily silent at the two endpoints.
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 will be able to identify a power series' centre, apply the ratio test with the variable carried through to get a condition on $|x - a|$ and hence the radius, say why the set of convergence must be an interval and why convergence inside it is absolute, test each endpoint separately as an ordinary numerical series, assemble an interval that may be closed at one end and open at the other, and recognise radii of zero and infinity as legitimate answers.
Every series so far was a list of numbers, and the question was whether it added up. $\sum x^n$ is not a list of numbers until $x$ is chosen — it is a whole family of series, one for each $x$, and different members behave differently. At $x = \tfrac12$ it converges, to $2$; at $x = 2$ it diverges.
So the question changes shape: not does it converge but for which $x$. The answer turns out to have a very rigid form — always an interval, always symmetric about the centre except possibly at the two endpoints — and the ratio test is what finds it.
A power series about $a$ is $\sum_{n\ge 0}c_n(x-a)^n$; the number $a$ is the centre, and the series always converges there, to $c_0$.
The radius of convergence $R$ is the number such that the series converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$. It may be $0$ (converges only at the centre) or infinite (converges everywhere).
The interval of convergence is the set of all $x$ where the series converges: the open interval of radius $R$, together with whichever endpoints happen to work. Finding $R$ is routine; the endpoints are a separate piece of work each.
Theorem. For any power series $\sum c_n(x-a)^n$ exactly one of three things is true: it converges only at $x = a$; it converges for every $x$; or there is an $R > 0$ such that it converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$.
That is a strong statement. The set of $x$ where a power series converges cannot be a scattered collection of points, or a union of two intervals — it is an interval centred at $a$, and the only freedom is at its two ends.
Finding $R$. Apply the ratio test to $|c_n(x-a)^n|$, keeping the $x$: $$\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right||x - a| < 1.$$ Solving for $|x-a|$ gives $R = \lim\left|\dfrac{c_n}{c_{n+1}}\right|$ when that limit exists. The root test gives $R$ from $\lim\sqrt[n]{|c_n|}$ and works in some cases the ratio test does not.
The endpoints. At $|x - a| = R$ the ratio test's limit is exactly $1$, so it is silent — necessarily, since both behaviours occur there. Substitute each endpoint and test the resulting series of numbers with the tools of the last six lessons: usually $p$-series, alternating, or divergence.
The two endpoints need not agree. $\sum\frac{x^n}{n}$ converges at $-1$ and diverges at $+1$. Four combinations are possible and all four occur.
Inside the radius the convergence is absolute, which is what makes the next lesson's term-by-term calculus safe.
Another way: picture
A segment of the number line centred at $a$, of half-width $R$, shaded for convergence. Everything strictly inside is shaded, everything strictly outside is blank, and the two endpoints are drawn as small circles whose filling is decided separately, one at a time. The name radius comes from the complex plane, where the picture is a genuine disc — and where the reason the region is round becomes visible.
Another way: steps
Nothing so far suggests that the set of good $x$ should be so well behaved. The reason is a small lemma with a large consequence.
If $\sum c_n b^n$ converges, then $\sum c_n x^n$ converges absolutely for every $|x| < |b|$. Because the terms $c_n b^n$ converge to zero they are bounded, say by $M$; then $$|c_n x^n| = |c_n b^n|\left|\frac{x}{b}\right|^n \le M\left|\frac{x}{b}\right|^n,$$ and the right-hand side is a geometric series with ratio below $1$. Comparison finishes it.
So convergence at any point forces absolute convergence at every point nearer the centre. The set of good $x$ is therefore "downward closed" in distance from $a$ — which is exactly what makes it an interval, and makes the radius its supremum.
This also explains the endpoints' special status. The lemma says nothing about points at the same distance, only nearer ones, so the boundary is genuinely undetermined by the interior. And it explains why the convergence inside is always absolute: the comparison was with a geometric series of absolute values.
In the complex plane the same argument gives a disc, and the name radius stops being metaphorical. It also explains something otherwise mysterious: $\dfrac{1}{1+x^2}$ is perfectly smooth on the whole real line, yet its series about $0$ has radius $1$. The obstruction is at $x = \pm i$, where the function blows up — invisible from the real line, and decisive.
Inferring the endpoints from the radius. The radius says nothing about them; that is the definition of an endpoint here.
Assuming the two endpoints behave alike. They frequently do not.
Forgetting the centre. $\sum \frac{(x-3)^n}{n}$ has radius $1$ about $3$, so the interval is around $3$ and not around $0$.
Applying the ratio test to $c_n$ alone. The $x$ is part of the term and has to be carried; without it there is no condition on $x$ to solve.
Reporting $R$ as the answer to "find the interval of convergence". The interval is the endpoints' verdicts as well.
Missing that the series may converge only at the centre. $\sum n!x^n$ has $R = 0$, and its interval of convergence is the single point $0$.
Losing a substitution's effect on the radius. A series in $x^2$ has its radius in $x$, not in $x^2$; take the square root.
"Find the interval of convergence" is answered, over and over, with a number. The radius is the easy half of the problem and the half that a routine calculation produces, so it feels like the result; the endpoints are fiddly and get dropped.
They are not a refinement. They are where all the interesting behaviour lives, and they are the reason the object is called an interval rather than a radius. $\sum\frac{x^n}{n}$, $\sum\frac{x^n}{n^2}$ and $\sum x^n$ all have radius $1$ about $0$ and have intervals $[-1, 1)$, $[-1,1]$ and $(-1,1)$ respectively. Three different answers from one radius.
And the ratio test cannot help, by necessity rather than by weakness. At $|x - a| = R$ its limit is exactly $1$, which is the value at which it is silent — and it must be silent, because all three of those series sit there and they do different things. A test that spoke at the boundary would be a test that gave the same answer to series with different answers.
So an interval-of-convergence problem is always three problems: one ratio test and two ordinary series tests. Budget for all three, and expect the two endpoints to disagree.
$\displaystyle\sum_{n=1}^{\infty}\frac{(x+1)^{n}}{n}$: centre $-1$. Ratio: $\dfrac{n}{n+1}|x+1| \to |x+1|$, so $R = 1$ and the open interval is $(-2, 0)$.
Ratio test with the $x$ carried through.
At $x = 0$: $\sum \dfrac1n$, the harmonic series, which diverges. Excluded.
One endpoint, one test.
At $x = -2$: $\sum \dfrac{(-1)^{n}}{n}$, which converges by Leibniz. Included. The interval is $[-2, 0)$ — closed at one end, open at the other.
$\displaystyle\sum_{n=0}^{\infty}n!\,x^{n}$: the ratio is $(n+1)|x|$, which tends to infinity for every $x \ne 0$. So the series diverges everywhere except at the centre: $R = 0$.
A legitimate radius.
$\displaystyle\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$: the ratio is $\dfrac{|x|}{n+1} \to 0$ for every $x$, so it converges everywhere: $R = \infty$, and there are no endpoints to test.
The other extreme.
The two are reciprocal in shape, and the second is $e^x$. A factorial in the denominator is what buys an infinite radius; one in the numerator destroys the series entirely.
Write it about its centre: $2x - 1 = 2\left(x - \tfrac12\right)$, so the centre is $\tfrac12$ and the term is $\dfrac{2^{n}(x - 1/2)^{n}}{n^{2}}$.
Find the centre before anything else.
Ratio: $\left(\dfrac{n}{n+1}\right)^{2}\cdot 2\left|x - \tfrac12\right| \to 2\left|x - \tfrac12\right|$, below $1$ when $\left|x - \tfrac12\right| < \tfrac12$. So $R = \tfrac12$ and the open interval is $(0, 1)$.
Solve for the distance from the centre.
Endpoints: at $x = 1$ the series is $\sum n^{-2}$ and at $x = 0$ it is $\sum(-1)^{n}n^{-2}$; both converge. The interval is $[0, 1]$.
Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}8^{\,n}x^{n}$.
Answer:
$\displaystyle\sum_{n=1}^{\infty}\frac{(x - 4)^{n}}{n\,3^{\,n}}$ has radius $3$ about $4$, converges at the left endpoint and diverges at the right. Give the interval of convergence.
This task has no paper form; do it on a device.
Does $\sum \dfrac{x^{n}}{\sqrt{n}}$ converge at $x = 1$?
Put the steps of finding the interval of convergence of $\displaystyle\sum\frac{(x - 3)^{n}}{n^{2}}$ into the order you do them.
Number the steps in order (write the number in the box):
Give the radius of convergence of each series about $0$.
| Radius | |
|---|---|
| $\sum \dfrac{x^{n}}{n!}$ | |
| $\sum 5^{\,n}x^{n}$ | |
| $\sum n!\,x^{n}$ |
$\displaystyle\sum_{n\ge 0}u^{n}$ has radius $1$. Substituting $u = 3x^{2}$ gives $\displaystyle\sum_{n\ge 0}3^{\,n}x^{2n}$. For which $|x|$ does that converge? Give the bound on $|x|$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}4^{\,n}x^{n}$.
Answer:
You can produce a full interval of convergence, endpoints included. Say in your own words why the ratio test must be silent at the endpoints rather than merely happening to be. Next: what you may do to a power series inside its radius.
10. Your turn: the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(2x - 1)^{n}}{n^{2}}$, step 3
Both ends included — a conclusion, not a default.