A power series
\sum c_n (x-a)^n converges for some values of
x and diverges for others. Remarkably, the convergence set is never
a ragged scatter — it is always an interval centred at
a. There is a number R \in [0, \infty],
the radius of convergence, such that the series
- converges (absolutely) whenever |x - a| < R, and
- diverges whenever |x - a| > R.
Only the two endpoints x = a \pm R are left undecided — each must
be checked by hand. The job of this page is to compute R,
and the tool is the
ratio test.
Deriving the radius from the ratio test
The ratio test looks at the limiting ratio of consecutive terms of a series
\sum u_n: if \lim |u_{n+1}/u_n| = L,
the series converges absolutely when L < 1 and diverges when
L > 1. Apply it to the power series with
u_n = c_n (x-a)^n.
Step 1 — form the ratio of consecutive terms.
\left|\frac{u_{n+1}}{u_n}\right| = \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n}\right| = \left|\frac{c_{n+1}}{c_n}\right|\,|x - a|.
Step 2 — take the limit. Suppose the coefficient ratio settles to a limit,
\displaystyle \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| = \ell.
Since |x-a| is a constant, it pulls outside the limit:
L = \lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right| = \ell \,|x - a|.
Step 3 — impose the convergence condition. The ratio test demands
L < 1 for convergence:
\ell\,|x - a| < 1 \quad\Longleftrightarrow\quad |x - a| < \frac{1}{\ell}.
Step 4 — read off the radius. Comparing with
|x - a| < R, the radius is the reciprocal of the coefficient ratio:
\boxed{\,R = \frac{1}{\ell} = \lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|\,}
with the natural conventions R = \infty when
\ell = 0 (the terms shrink so fast the series converges
everywhere) and R = 0 when
\ell = \infty (it converges only at the centre).
Three worked radii
(a) \sum x^n/n! gives R = \infty.
Here c_n = 1/n!, so
\left|\frac{c_{n+1}}{c_n}\right| = \frac{n!}{(n+1)!} = \frac{1}{n+1} \to 0 = \ell \quad\Longrightarrow\quad R = \frac{1}{0} = \infty.
Converges for every x — this is the series for
e^x.
(b) \sum n!\,x^n gives R = 0.
Now c_n = n!, so
\left|\frac{c_{n+1}}{c_n}\right| = \frac{(n+1)!}{n!} = n+1 \to \infty = \ell \quad\Longrightarrow\quad R = \frac{1}{\infty} = 0.
Converges only at x = 0 — a power series that is barely a function
at all.
(c) \sum x^n/n gives R = 1,
endpoints differing. Here c_n = 1/n:
\left|\frac{c_{n+1}}{c_n}\right| = \frac{n}{n+1} \to 1 = \ell \quad\Longrightarrow\quad R = 1.
So it converges on (-1, 1) and diverges outside
[-1, 1]. The endpoints must be tested separately. At
x = 1 the series is \sum 1/n, the
harmonic series — diverges. At x = -1 it is
\sum (-1)^n/n, the alternating harmonic series —
converges. The interval of convergence is therefore
[-1, 1): same radius, different fates at the two ends.
For a power series \sum c_n (x-a)^n there is a radius
R \in [0, \infty] such that:
-
it converges absolutely for |x - a| < R and diverges for
|x - a| > R;
-
when the limit exists,
R = \lim_{n\to\infty}\left|\dfrac{c_n}{c_{n+1}}\right| (ratio
test), or more generally
1/R = \limsup_{n\to\infty}\sqrt[n]{|c_n|} (Cauchy–Hadamard,
root test);
-
the two endpoints x = a \pm R are not decided
by R and must be checked individually — convergence there can
go either way.
The three worked examples are the three archetypes, and they are genuinely different
species of object:
-
R = 0 (\sum n! x^n): the series is a
function on a single point. Useful only as a formal power series — an algebraic
bookkeeping device, not an analytic function.
-
R finite (\sum x^n/n): a real
function on a bounded interval, with the interesting drama all at the endpoints.
-
R = \infty (\sum x^n/n!): an
entire function, defined on all of
\mathbb{R} (indeed all of \mathbb{C}).
The exponential, sine and cosine all live here.
The endpoint subtlety in case (b) of the worked example —
[-1,1) — is not a blemish but the whole story: the radius is a
blunt instrument that locates the boundary, and the fine question of what happens
on the boundary needs the convergence tests for ordinary numerical series,
including the alternating-series test that rescued x = -1.