JOVANA
Explore Library Glossary Getting Started Three Levels Fields How it works Mission
Join the mission
All guides

Series of Complex Numbers and Convergence

Before we can write a holomorphic function as an infinite power series, we need to know what it even means to add up infinitely many complex numbers. We build convergence from the ground up, meet absolute convergence as the well-behaved kind, and pick up the two tests that will hand us radii of convergence in the next guide.

What it means to add up infinitely many complex numbers

You climbed this far knowing how to add, multiply, and take limits of complex numbers, and you have seen sequences in the plane settle down toward a point. A series is nothing more exotic than a sequence in disguise. Given numbers a_0, a_1, a_2, ..., we form the partial sums s_N = a_0 + a_1 + ... + a_N, and we say the series sum a_n converges to a number s exactly when the sequence s_0, s_1, s_2, ... of these running totals converges to s. So a series is just a sequence (of partial sums) wearing a fancier hat.

Because a complex number z = x + i y lives in the plane, this convergence splits cleanly: s_N converges if and only if its real part and its imaginary part each converge as ordinary real series. Picture the partial sums as dots hopping around the plane; convergence means the dots eventually crowd into a tiny disk around s and never escape it. This is the same as saying the modulus |s_N - s| shrinks to zero. Everything you already trust about real series — that the tail must go to zero, that you can add convergent series term by term — carries over coordinate by coordinate.

Absolute convergence: the well-behaved kind

There is a stronger, friendlier notion. A series sum a_n is absolutely convergent when the real series of moduli, sum |a_n|, converges. The headline fact is that absolute convergence forces ordinary convergence: if sum |a_n| converges, then sum a_n converges too. The reason is short and worth keeping in your pocket — the partial sums of sum a_n form a Cauchy sequence because, by the triangle inequality, the length of any block of terms |a_{m+1} + ... + a_n| is no bigger than |a_{m+1}| + ... + |a_n|, and that tail is already known to be tiny.

Why care about the distinction between absolute and conditional convergence? Because absolutely convergent series are robust in ways ordinary ones are not: you may reorder their terms freely, regroup them, and even multiply two of them together (the Cauchy product) without breaking convergence or changing the sum. A series that converges but not absolutely is called conditionally convergent, and rearranging it can change its value — so it is delicate, and we mostly steer around it. The good news for power series is that, inside their disk of convergence, they converge absolutely; the delicacy lives only on the boundary circle.

Two tests that compare against a geometric series

Almost every convergence test we will use measures a series against the one series we understand completely: the geometric series sum r^n. For a complex ratio r, that series converges exactly when |r| < 1 (then its sum is 1/(1 - r)) and diverges when |r| >= 1. So the strategy of the ratio and root tests is to ask whether, deep in the tail, the terms shrink like the powers of some r with |r| < 1.

Ratio test:  L = lim |a_(n+1) / a_n|        (n -> infinity)
Root test:   L = lim sup |a_n|^(1/n)

   L < 1  ->  converges absolutely
   L > 1  ->  diverges
   L = 1  ->  test is silent (inconclusive)
Both tests compare the size of the terms against a geometric benchmark; the root test is the stronger and is what makes the radius-of-convergence formula work.

A small but honest point: when L = 1 both tests shrug, and you must look harder by some other means. Also, the root test (using the lim sup of |a_n|^(1/n)) is genuinely stronger than the ratio test — it works on more series and never gives a smaller verdict — which is exactly why the Cauchy-Hadamard formula for a radius of convergence is built from roots rather than ratios. The ratio test is just the friendlier one to compute when the terms have clean factorials or powers.

A worked example: the exponential series

Let us put the machinery to work on the most important series in the subject, the one that defines the complex exponential e^z = sum z^n / n!. Fix any complex number z, treat n as the running index, and apply the ratio test to the terms a_n = z^n / n!.

  1. Form the ratio of consecutive terms: a_(n+1) / a_n = (z^(n+1) / (n+1)!) divided by (z^n / n!), which simplifies to z / (n + 1).
  2. Take moduli: |a_(n+1) / a_n| = |z| / (n + 1). The fixed number |z| sits on top, while the bottom grows without bound.
  3. Let n go to infinity: the limit L = lim |z| / (n + 1) = 0, no matter which z you started with.
  4. Since L = 0 < 1 for every z, the series converges absolutely everywhere in the plane — so e^z is defined on all of the complex numbers.

That single computation is quietly enormous. Because the exponential series converges everywhere, e^z is a function with no points of breakdown — the kind we will later call entire. The same series, restricted to z = i theta, is exactly the engine behind Euler's formula e^(i theta) = cos theta + i sin theta, which you have been using since the polar-form rung. So the tidy convergence we just checked is the rigorous foundation under a formula you already trust.

Why this matters: the road to power series

Everything in this guide is the warm-up for a much bigger idea. A power series is a series whose terms carry a variable: sum a_n (z - z_0)^n. For a fixed z it is just a series of complex numbers like the ones above, so we can ask the same convergence questions — and the answer turns out to be beautifully geometric. There is a number R, the radius of convergence, so that the series converges (absolutely) when |z - z_0| < R and diverges when |z - z_0| > R. The next guide builds that disk and shows how the root test pins down R.