Radius and Interval of Convergence of Power Series

AP Calculus BC · Infinite Sequences and Series

Radius and Interval of Convergence

The ratio test settles the interior in one line; the endpoints each need their own.

key idea$\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert\lt 1\ \Rightarrow\ \lvert x-c\rvert\lt R$

The procedure

A power series converges on an interval centred at $c$, and the ratio test finds it. The test settles the interior in one computation and is always inconclusive at the two endpoints — which must therefore be checked separately, one at a time.

Apply the ratio test to $\lvert a_n\rvert$

Form $\displaystyle\lim_{n\to\infty}\left\lvert\dfrac{a_{n+1}}{a_n}\right\rvert$. The factor $\lvert x-c\rvert$ pulls out of the limit as a constant.

in the series out a limit of the form $\lvert x-c\rvert\cdot L$

Set the limit less than 1

$\lvert x-c\rvert L\lt1$ gives $\lvert x-c\rvert\lt\dfrac1L$, so the radius is $R=\dfrac1L$.

out the radius $R$

Write the open interval

$c-R\lt x\lt c+R$. This much is guaranteed; the endpoints are not yet decided.

out $(c-R,\,c+R)$

Test each endpoint separately

Substitute $x=c-R$ and then $x=c+R$ into the original series. Each substitution gives a numerical series needing its own test — often alternating on one side and a $p$-series on the other.

out two independent verdicts

Assemble the interval

Use a square bracket where the endpoint converges and a round one where it does not. The two ends are decided independently and need not match.

out the interval of convergence

Power series$\displaystyle\sum a_n(x-c)^{n}$ — a series whose terms carry powers of $x-c$.

Centre $c$The point the interval is symmetric about. Read it from $(x-c)$, minding the sign.

Radius $R$Half the interval's width. $R=0$ means convergence only at $c$; $R=\infty$ means everywhere.

Interval of convergenceThe full set of $x$ for which the series converges, including whichever endpoints survive their own test.

Endpoint testA separate convergence test on the numerical series obtained by substituting that endpoint.

Series	$R$	Interval
$\displaystyle\sum\frac{x^{n}}{n!}$	$\infty$	$(-\infty,\infty)$
$\displaystyle\sum n!\,x^{n}$	$0$	$\{0\}$ only
$\displaystyle\sum x^{n}$	$1$	$(-1,1)$ — both ends diverge
$\displaystyle\sum\frac{x^{n}}{n}$	$1$	$[-1,1)$ — only the left end survives
$\displaystyle\sum\frac{x^{n}}{n^{2}}$	$1$	$[-1,1]$ — both ends converge

The last three all have $R=1$ and three different intervals. The radius never determines the endpoints — that is exactly why they get their own tests.

Worked examples

EX 1

One endpoint survives

the classic asymmetric case

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\dfrac{x^{n}}{n}$.

Apply the ratio test

$\left\lvert\dfrac{x^{n+1}/(n+1)}{x^{n}/n}\right\rvert=\lvert x\rvert\cdot\dfrac{n}{n+1}\to\lvert x\rvert$.

Set it below 1

$\lvert x\rvert\lt1$, so $R=1$ and the open interval is $(-1,1)$.

Test $x=1$

$\displaystyle\sum\dfrac1n$ — the harmonic series, which diverges.

Test $x=-1$

$\displaystyle\sum\dfrac{(-1)^{n}}{n}$ — converges by the alternating series test.

Assemble

Interval of convergence: $[-1,1)$.

EX 2

Both endpoints converge

a squared denominator

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\dfrac{(x-3)^{n}}{n^{2}}$.

Read the centre

$(x-3)$, so $c=3$.

Apply the ratio test

$\lvert x-3\rvert\cdot\dfrac{n^{2}}{(n+1)^{2}}\to\lvert x-3\rvert$, so $R=1$.

Write the open interval

$2\lt x\lt4$.

Test both endpoints

At $x=4$: $\displaystyle\sum\dfrac{1}{n^{2}}$ converges ($p=2$). At $x=2$: $\displaystyle\sum\dfrac{(-1)^{n}}{n^{2}}$ converges absolutely.

Assemble

$[2,4]$ — both endpoints included.

EX 3

Infinite radius

factorials in the denominator

Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n}}{n!}$.

Apply the ratio test

$\left\lvert\dfrac{x^{n+1}/(n+1)!}{x^{n}/n!}\right\rvert=\dfrac{\lvert x\rvert}{n+1}$.

Take the limit

$\displaystyle\lim_{n\to\infty}\dfrac{\lvert x\rvert}{n+1}=0$ for every $x$.

Compare with 1

$0\lt1$ always, so the series converges for all real $x$.

State the radius

$R=\infty$.

Note the absence of endpoints

There are none to test. This series is the Maclaurin series for $e^{x}$, which is why it converges everywhere.

EX 4

Zero radius

factorials in the numerator

Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty}n!\,(x-2)^{n}$.

Apply the ratio test

$\left\lvert\dfrac{(n+1)!\,(x-2)^{n+1}}{n!\,(x-2)^{n}}\right\rvert=(n+1)\lvert x-2\rvert$.

Take the limit

For any $x\neq2$ this tends to $\infty$.

Apply the criterion

A limit above $1$ means divergence, so the series fails for every $x\neq2$.

Check the centre

At $x=2$ every term after the first is $0$, so the series trivially converges.

State the result

$R=0$; the series converges only at $x=2$. A power series always converges at its centre — sometimes only there.

Where it goes wrong

Endpoints always need their own tests

The ratio test returns exactly $1$ there and is inconclusive by construction. Substitute each endpoint and test the resulting numerical series.

The two ends are independent

$[-1,1)$ is a perfectly normal answer. Nothing requires the interval to be symmetric in its bracket types.

Read the centre from $(x-c)$

$(x+4)^{n}$ has centre $c=-4$. The sign inside the parentheses is the opposite of the centre.

$R=0$ and $R=\infty$ are real answers

Factorials in the numerator collapse the interval to a point; factorials in the denominator open it to the whole line.

Course map. Topic 10.13 · LO LIM-8.D — determine the radius and interval of convergence of a power series. BC-only.