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A series with an $x$ in it is a family of series, one for each $x$; the ratio test with the $x$ carried through gives the radius, and the set of good $x$ is always an interval about the centre.
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 read the centre of a power series off its bracket, apply the ratio test with the variable carried through to get a condition on $|x - a|$, and solve that condition for the radius of convergence. You can write the open interval the radius gives, recognise $0$ and infinity as legitimate radii, and say why a radius is not yet an answer to a question about an interval.
You can apply the ratio test: form the size of the ratio of consecutive terms, take its limit, and compare with $1$. Every series you have tested so far was a list of numbers. This section puts a letter in one, and the ratio test turns out to answer the new question almost unchanged.
Power series about $a$: $\sum_{n\ge 0}c_n(x - a)^n$.
Centre: the number $a$; the series always converges there, to $c_0$.
Radius of convergence $R$: the series converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$.
Interval of convergence: every $x$ at which the series converges — the open interval of radius $R$, plus whichever endpoints happen to work.
$\sum x^n$ is not a series until $x$ is chosen. At $x = \tfrac12$ it converges, to $2$; at $x = 2$ it diverges. So it is a whole family of series, and the question changes shape: not does it converge but for which $x$.
The answer always has the same rigid form. 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 a number $R > 0$ with
$$|x - a| < R \implies \text{converges}, \qquad |x - a| > R \implies \text{diverges}.$$
The set of good $x$ cannot be scattered points or two separate pieces. 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 $x$. That gives a condition of the form $L\,|x - a| < 1$, and solving it for $|x - a|$ gives $R$. Keeping the $x$ is the whole move: drop it and the ratio is a statement about the coefficients with no $x$ in it to solve for.
$0$ and infinity are radii. $\sum n!\,x^n$ has $R = 0$ and converges only at its centre; $\sum \frac{x^n}{n!}$ has $R = \infty$ and converges everywhere. Both are legitimate answers rather than signs of an error.
The centre is not always zero. $\sum \frac{(x - 3)^n}{n}$ has radius $1$ about $3$, so its interval sits around $3$ and not around the origin.
Another way: picture
Draw a segment of the number line centred at $a$, reaching $R$ in each direction, and shade it. Everything strictly inside is shaded for convergence, everything strictly outside is blank, and the two endpoints are small circles whose filling is decided separately, one at a time. The word radius comes from the complex plane, where the same picture is a genuine disc about the centre.
Another way: steps
| Series | Ratio of consecutive terms | Radius |
|---|---|---|
| $\sum x^n$ | $\lvert x\rvert$ | $1$ |
| $\sum \dfrac{x^n}{n}$ | $\dfrac{n}{n+1}\lvert x\rvert \to \lvert x\rvert$ | $1$ |
| $\sum 3^n x^n$ | $3\lvert x\rvert$ | $\tfrac13$ |
| $\sum \dfrac{x^n}{n!}$ | $\dfrac{\lvert x\rvert}{n+1} \to 0$ | infinite |
| $\sum n!\,x^n$ | $(n+1)\lvert x\rvert \to \infty$ | $0$ |
A polynomial factor in the coefficients — an $n$, an $n^2$, a $\sqrt{n}$ — makes no difference to the radius at all, because $\frac{n}{n+1}$ and its relatives all tend to $1$. Only a geometric factor or a factorial moves it.
Dropping the $x$ from the ratio. The $x$ is part of the term. Without it the ratio test gives a number rather than a condition, and there is nothing left to solve for.
Measuring from zero instead of from the centre. Radius $2$ about the centre $5$ means $|x - 5| < 2$, which is the interval from $3$ to $7$ — not from $-2$ to $2$.
Reporting the radius as the interval. They are different objects: one is a number, the other a set with two ends. The radius is the routine half of the work, which is exactly why it gets mistaken for the whole of it.
$\sum 4^n x^n$: the ratio of consecutive terms in size is $\dfrac{4^{n+1}|x|^{n+1}}{4^n |x|^n} = 4|x|$.
Keep the $x$.
There is no $n$ left, so the limit is $4|x|$, and the ratio test wants that below $1$.
Set it below one.
$|x| < \tfrac14$, so $R = \tfrac14$. The coefficients grow by a factor of $4$ each step, and the radius is the reciprocal of that.
$\sum \dfrac{(x - 5)^n}{n^2}$: the centre is $5$, read straight off the bracket.
Find the centre first.
The ratio is $\left(\dfrac{n}{n+1}\right)^2 |x - 5| \to |x - 5|$, so the condition is $|x - 5| < 1$ and $R = 1$.
The $n^2$ changes nothing.
The open interval runs from $4$ to $6$. Reporting $(-1, 1)$ here is the standard slip, and it comes from solving $|x| < 1$ instead.
The ratio of consecutive terms in size is $\dfrac{n}{n+1}\cdot 2|x|$, which tends to $2|x|$.
Keep the $x$ and let the $n$ go.
Set that below $1$: $2|x| < 1$, so $|x| < \tfrac12$.
Solve for the distance from the centre.
So $R = \tfrac12$ about the centre $0$, and the open interval runs from $-\tfrac12$ to $\tfrac12$ — with both ends still to be tested.
Put the steps of finding the radius of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x - 5)^{n}}{n}$ into the order you do them.
Number the steps in order (write the number in the box):
Give the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}5^{\,n}x^{n}$, then of $\displaystyle\sum_{n=0}^{\infty}\frac{x^{n}}{5^{\,n}}$.
first radius = p, second radius = q
A power series is centred at $-4$ and has radius of convergence $4$. Mark the right-hand end of its interval of convergence.
-12 |——————————| 12
Mark the position with a cross, then write the value:
Give the radius of convergence of each series about $0$.
| Radius | |
|---|---|
| $\sum \dfrac{x^{n}}{n!}$ | |
| $\sum 3^{\,n}x^{n}$ | |
| $\sum n!\,x^{n}$ |
A student is asked for the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x - 3)^{n}}{n}$ and answers "the radius is $1$". What is still missing?
$\displaystyle\sum_{n\ge 0}u^{n}$ converges exactly when $|u| < 1$. Substituting $u = 3x^{2}$ turns it into $\displaystyle\sum_{n\ge 0}3^{\,n}x^{2n}$. For which values of $x^{2}$ does that converge? Give the bound on $x^{2}$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of finding the radius of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x - 5)^{n}}{n}$ into the order you do them.
Number the steps in order (write the number in the box):
You can find the radius of convergence of a power series. Without looking: what has to stay in the ratio for the test to say anything about $x$, and what does a radius of $2$ about the centre $5$ mean?
9. Your turn: the radius of $\displaystyle\sum_{n=1}^{\infty}\frac{(2x)^{n}}{n}$, step 3