Radius of Convergence
When a calculator approximates a function by adding up a power series, the trick only works
inside a certain window around the centre; step outside it and the sum turns to nonsense.
Knowing exactly how wide that trustworthy window is — its radius of
convergence — is what keeps every such approximation honest.
A power series
\sum c_n (x-a)^n is an infinite polynomial, and an infinite
polynomial is a gamble: plug in an x and the sum might settle to a
number, or it might explode. You would expect the winning values of
x to be a mess — a scatter of lucky points, different for every
series. They never are. Every power series comes with a blast radius: a single
number R \in [0, \infty] such that
-
inside — whenever |x - a| < R — the series
converges, and not grudgingly but absolutely: rearrange it, differentiate it,
integrate it term by term, and everything behaves beautifully;
-
outside — whenever |x - a| > R — the series
diverges, without exception: the terms don't even shrink to zero.
No half-measures, no scattered islands of convergence out beyond the boundary. The only
points the number R refuses to classify are the two boundary points
x = a \pm R themselves, and each of those must be interrogated by
hand. That is the entire theory of where a power series lives, compressed into one number — and
this page is about how to compute that number. The workhorse tool is the
ratio test.
For every power series \sum c_n (x-a)^n there is a unique
R \in [0, \infty], the radius of convergence, and
exactly one of three things happens:
-
R = 0 — the series converges only at its centre
x = a;
-
0 < R < \infty — it converges absolutely for
|x - a| < R and diverges for
|x - a| > R; the endpoints
x = a \pm R are undecided and must be tested individually;
-
R = \infty — it converges absolutely for every
x.
-
When the limit exists,
R = \lim_{n\to\infty}\left|\dfrac{c_n}{c_{n+1}}\right| (via the
ratio test);
-
in full generality,
\dfrac{1}{R} = \limsup_{n\to\infty}\sqrt[n]{|c_n|}
(Cauchy–Hadamard, via the root 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 — treating x as a
fixed but unknown number — and the radius falls straight out.
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 as far as n is
concerned, 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 limit:
\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 coefficients shrink so fast the series converges
everywhere) and R = 0 when \ell = \infty
(it converges only at the centre). Notice what happened to x in
Step 3: the condition L < 1 carved the number line into an
inside and an outside — which is exactly why the convergence set is always an
interval and never a ragged scatter.
Three radii, three worlds
(a) \sum x^n/n! gives R = \infty.
Here c_n = 1/n!, so the coefficient ratio is
\left|\frac{c_{n+1}}{c_n}\right| = \frac{n!}{(n+1)!} = \frac{1}{n+1} \;\longrightarrow\; 0 = \ell \quad\Longrightarrow\quad R = \frac{1}{0} = \infty.
The factorial in the denominator crushes the powers of x no matter
how large |x| is — eventually n! outgrows
|x|^n and the terms plummet. The series converges for every real
x: no endpoints to check, because there is no boundary. This is the
series for e^x.
(b) \sum n!\,x^n gives R = 0.
Flip the factorial upstairs: now c_n = n!, so
\left|\frac{c_{n+1}}{c_n}\right| = \frac{(n+1)!}{n!} = n+1 \;\longrightarrow\; \infty = \ell \quad\Longrightarrow\quad R = \frac{1}{\infty} = 0.
However tiny you make x \neq 0, the factorial eventually overwhelms
it and the terms blow up. The series converges only at the single point
x = 0 — a power series that is barely a function at all. Again no
endpoint check: a point has no endpoints.
(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} \;\longrightarrow\; 1 = \ell \quad\Longrightarrow\quad R = 1.
So it converges on (-1, 1) and diverges outside
[-1, 1] — and now, for the first time, there is a genuine boundary
to interrogate. 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.
The full ritual, start to finish
Exam questions rarely centre a series at 0, and they always want the
endpoints. Here is the complete procedure on a typical specimen:
\sum_{n=1}^{\infty} \frac{(x-3)^n}{n \cdot 2^n}.
Step 1 — ratio test for the radius. With
u_n = \dfrac{(x-3)^n}{n\,2^n},
\left|\frac{u_{n+1}}{u_n}\right| = \frac{|x-3|^{n+1}}{(n+1)\,2^{n+1}} \cdot \frac{n\,2^n}{|x-3|^n} = \frac{n}{n+1}\cdot\frac{|x-3|}{2} \;\longrightarrow\; \frac{|x-3|}{2}.
Convergence requires \dfrac{|x-3|}{2} < 1, i.e.
|x - 3| < 2. So R = 2, the centre is
a = 3, and the interior of convergence is the interval from
3 - 2 = 1 to 3 + 2 = 5. The candidates
still on trial: the two endpoints x = 1 and
x = 5.
Step 2 — test the right endpoint x = 5. Substitute
it in — the powers collapse into pure numbers:
\sum_{n=1}^{\infty} \frac{(5-3)^n}{n\,2^n} = \sum_{n=1}^{\infty} \frac{2^n}{n\,2^n} = \sum_{n=1}^{\infty} \frac{1}{n}.
The harmonic series — diverges. x = 5 is out.
Step 3 — test the left endpoint x = 1.
\sum_{n=1}^{\infty} \frac{(1-3)^n}{n\,2^n} = \sum_{n=1}^{\infty} \frac{(-2)^n}{n\,2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}.
The alternating harmonic series — converges (by the alternating series test:
terms decreasing to zero). x = 1 is in.
Verdict. Radius R = 2, interval of convergence
\boxed{\,[1, 5)\,}
— closed at the left, open at the right. Notice that the two endpoint series were
ordinary numerical series, and each needed its own classical test. That is the part
everyone skips, and the part every examiner checks.
Two mistakes account for most lost marks on this topic:
-
The ratio test never settles the endpoints. At
|x - a| = R the ratio-test limit is
L = \ell \cdot R = 1 exactly — the one value where the
test is silent, by design. You must substitute each endpoint and run a separate
test on the resulting numerical series (comparison, p-series,
alternating series, divergence test…). And all four outcomes genuinely occur, all with
R = 1: \sum x^n keeps
neither endpoint, giving (-1,1);
\sum x^n/n^2 keeps both, giving
[-1,1]; \sum x^n/n keeps only the
left, giving [-1,1); and
\sum (-1)^n x^n/n keeps only the right, giving
(-1,1]. No shortcut predicts which — you check.
-
R is measured from the centre
a, not from 0. The series
\sum (x-7)^n has R = 1, and it
converges on (6, 8) — nowhere near the origin. In particular it
diverges at x = 0, since
|0 - 7| = 7 > 1. Always write the interval as
(a - R,\; a + R) before touching the endpoints.
The trichotomy is not just bookkeeping — the three cases produce genuinely different kinds of
mathematical object:
-
R = 0 (\sum n!\, x^n): a function on
a single point. Useless for calculus, but not useless — it survives as a formal
power series, an algebraic bookkeeping device where the coefficients carry the information
(combinatorics runs on these).
-
R finite (\sum x^n/n): a real
function on a bounded interval, with all the interesting drama at the two endpoints. In
fact \sum x^n/n = -\ln(1-x) on its interval — and the endpoint
that converges, x=-1, quietly evaluates the alternating harmonic
series to \ln 2.
-
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 — which is why you may plug anything into their series without a
moment's anxiety.
Take the geometric series \dfrac{1}{1-x} = \sum x^n, with
R = 1. Fair enough: the function itself blows up at
x = 1, so of course the series must stop there. The radius is the
distance from the centre to the nearest disaster. Sensible.
Now try \dfrac{1}{1+x^2} = \sum (-1)^n x^{2n}. Same radius,
R = 1 — but this function is perfectly smooth, bounded, and
infinitely differentiable on the entire real line. There is no disaster anywhere in
sight. Why on earth does its series give up at |x| = 1?
Because the disaster is hiding off the page. Allow x to be a
complex number and 1 + x^2 = 0 at
x = \pm i — two poles sitting one unit above and below the real
axis, invisible to real eyes but exactly distance 1 from the
centre 0. The deep theorem of complex analysis:
the radius of convergence equals the distance from the centre to the nearest singularity
in the complex plane. In \mathbb{C} the convergence set is a
genuine disc — the word "radius" is literal, not a metaphor. Our real interval
(a-R,\, a+R) is just the slice of that disc lying along the real
axis. And the theorem even makes predictions: recentre the same function at
a = 3 and the nearest pole \pm i is at
distance \sqrt{3^2 + 1^2} = \sqrt{10}, so that Taylor series has
R = \sqrt{10} — no ratio test required.
See the interval shrink and grow
Consider the family \sum n^{p}\, x^n, with a slider for the exponent
p. The coefficient ratio
|c_n/c_{n+1}| = (n/(n+1))^{p} \to 1 for every
p, so the radius is always R = 1: the
curve below (a partial sum, the first 18 terms) settles calmly inside
(-1, 1) and blows up outside it no matter where you put the slider.
What p does change is the endpoint story — the one thing
the radius cannot see. At p = -2 the coefficients are
1/n^2 and both endpoints converge (interval
[-1,1]); at p = -1 you get our
[-1, 1); at p = 0 the geometric series
keeps neither endpoint; and for p > 0 the terms don't
even tend to zero at x = \pm 1, so divergence there is instant.
Watch how much more violently the partial sum rears up near x = \pm 1
as you push p upward: same radius, ever-worse behaviour on the
boundary.
See it explained