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Power Series and the Radius of Convergence

A power series in the complex plane never converges on a strange, lopsided set: it locks onto a perfect disk. This guide builds that disk, names its radius, hands you a formula that computes it from the coefficients alone, and shows why inside the disk the series behaves as beautifully as any function ever could.

From a series of numbers to a series of functions

In the previous guide you learned what it means for a series of complex numbers to converge, and you met the comforting fact that absolute convergence implies convergence. Now we let the terms move. A power series is an expression sum a_n (z - z_0)^n, where the coefficients a_n and the centre z_0 are fixed complex numbers but z is a free variable. For each value of z you plug in, you get an ordinary number series — one that might converge or might blow up. The question that organizes everything is simply: for which z does it converge?

The model to keep in your head is the geometric series sum z^n. It converges exactly when |z| < 1, to the tidy value 1/(1 - z), and diverges the instant |z| >= 1. Notice the shape of the answer: not a strip, not a square, but the open unit disk — everything within distance 1 of the centre 0. That circular shape is not a quirk of this one example. It is the universal verdict for every power series, and our whole job is to understand why.

Why the region of convergence is always a disk

The engine is one small observation, sometimes called Abel's lemma. Suppose the series converges at some single point z_1. Then its terms a_n (z_1 - z_0)^n must shrink to zero, so in particular they are bounded — say all of them have modulus at most M. Now take any z strictly closer to the centre than z_1 is, and write r = |z - z_0| / |z_1 - z_0| < 1. Then |a_n (z - z_0)^n| is at most M r^n, and sum M r^n is just a geometric series with ratio r < 1. By comparison, the series at z converges absolutely.

Read that consequence carefully, because it is the whole secret. The moment a power series converges at one point, it converges at every point nearer the centre. Convergence cannot have holes and cannot prefer one direction over another — it only cares about distance |z - z_0|. So the set of good z is determined by a single number R: the supremum of all distances at which convergence happens. Inside the circle of radius R the series converges; outside it diverges. That number R is the radius of convergence, and the open region it carves out is the disk of convergence.

Computing R from the coefficients

Knowing R exists is satisfying, but you also want to compute it. The cleanest tool reuses the ratio and root tests from the previous guide. The root test says sum c_n converges absolutely when limsup |c_n|^(1/n) < 1. Apply it with c_n = a_n (z - z_0)^n: you get limsup |a_n|^(1/n) times |z - z_0| < 1. That inequality is exactly |z - z_0| < 1 / limsup |a_n|^(1/n), which hands you R directly.

1 / R  =  limsup  |a_n|^(1/n)       (Cauchy-Hadamard)
              n -> infinity

  R = infinity  when the limsup is 0   (e.g. a_n = 1/n!)
  R = 0         when the limsup is +inf (e.g. a_n = n!)

quick shortcut, IF the limit exists:

  R = lim |a_n / a_(n+1)|         (ratio form)
      n -> infinity
The Cauchy-Hadamard formula always works; the ratio shortcut only when that limit exists.

This is the Cauchy-Hadamard formula, and the word limsup in it is doing real work. The ordinary limit of |a_n|^(1/n) need not exist — coefficients can oscillate — but the limit superior always does, so the formula never fails to define R. The familiar ratio test, R = lim |a_n / a_(n+1)|, is a convenient shortcut that you may use only when that limit genuinely exists. For a_n = 1/n! you get R = infinity, which is why e^z converges in the whole plane; for a_n = n! you get R = 0, a series that converges nowhere but its centre.

  1. Read off the coefficient a_n as a formula in n (be careful about gaps — a missing power means that a_n is 0).
  2. Try the ratio form R = lim |a_n / a_(n+1)| first; it is usually the least painful arithmetic.
  3. If that limit refuses to exist (oscillating or gappy coefficients), fall back to 1/R = limsup |a_n|^(1/n).
  4. State the disk |z - z_0| < R, and remember you have proved nothing yet about the boundary circle.

Inside the disk, the series is as good as gold

Why do we care so much about the disk? Because everywhere strictly inside it the convergence is not merely pointwise but uniform on every closed sub-disk — this is uniform convergence on compacta. Pick any radius r < R; on the closed disk |z - z_0| <= r the partial sums march to their limit at a single shared rate, with the geometric tail M r'^n controlling them all at once (here r' = r/R-style ratio is below 1). Uniform convergence is the magic ingredient that lets you exchange limits with sums and integrals without fear.

The headline consequence is term-by-term differentiation. Inside its disk a power series defines a holomorphic function f(z), and its derivative is obtained simply by differentiating each term: f'(z) = sum n a_n (z - z_0)^(n-1). Crucially, the differentiated series has the same radius R — the factor n out front does not change limsup |n a_n|^(1/n), since n^(1/n) tends to 1. So you can differentiate again, and again, forever. A power series is automatically infinitely differentiable inside its disk, with no extra hypotheses.

What the radius is really measuring

Here is the picture that makes R intuitive rather than a formula to memorize. The series sum z^n stops at |z| = 1 not because the algebra runs out, but because its sum 1/(1 - z) misbehaves at z = 1: the function it builds has a pole sitting on that circle. This is no coincidence. For any power series, R is exactly the distance from the centre to the nearest point where the function it represents stops being holomorphic — its nearest singularity. The disk is the largest one you could draw around z_0 that hits no trouble.

This explains something that looks paradoxical from the real line. The real function 1/(1 + x^2) is smooth and bounded for every real x, yet its Taylor series at 0 only converges for |x| < 1. From the real viewpoint there is no reason for it to stop. From the complex viewpoint the reason is obvious: 1/(1 + z^2) has poles at z = i and z = -i, each at distance 1 from the origin, and the radius simply reaches out to touch them. The complex plane sees obstructions that the real line cannot.

So the radius of convergence is never an arbitrary number bolted onto a series — it is the function quietly reporting where its troubles lie. A later guide in this rung will make that claim airtight; for now, let it reshape how you read every series you meet. When you compute R, you are locating the closest crack in the function's perfection, measured straight out from the point you expanded around.