Power series are most important in the context of

complex analysis, and most of the time the term power series is used to refer to a complex power series (

real power series are best considered as a special case of complex power series).

The first important property of power series is that they converge on some disc, so the define a function on a nice and sensible domain. The idea of the proof is quite simple, but it uses extra (simple) machinery of uniform convergence to show that the function is not just well-defined on the disc but also continuous.

The second important property is that a function defined by a power series is in fact not only continuous but also complex differentiable. Thus power series give analytic functions, which are extremely nice objects. There is also a converse to this: Taylor's theorem for complex functions states that any analytic function can be expressed locally as a power series (note that this does not quite hold for real functions, even if they are infinitely differentiable)

**Proposition**:

Given a sequence of complex numbers a_{n}, n = 1, 2, 3, ..., there is a non-negative real number R (which may take the values 0 and ∞), called the radius of convergence, such that the series Σ a_{n}z^{n} (this and all subsequent sums are taken from n = 0 to ∞) converges for all z in the disc A = {z ∈ **C** : |z| < R}.
The series converges locally uniformly on A, and the function defined by the series is continuous on A.

**Proof**:

Suppose that the Σ a_{n}w^{n} converges for some w ∈ **C**. Then the sequence a_{n}w^{n} must be bounded by some constant M. Let k be any real number < 1, and B the disc {z ∈ **C** : |z| < k|w|}. Then

Σ sup_{z∈E}|a_{n}z^{n}| < Σ |a_{n}k^{n}w^{n}|
< Σ Bk^{n} = M/(1-k)

so by the Weierstrass M-test Σ a_{n}z^{n} converges uniformly on B.

Thus if we let R = sup {|z| : Σ a_{n}z^{n} converges} we have that Σ a_{n}z^{n} converges uniformly on the disc |z| < r for any r < R. Hence for any z ∈ A we have that Σ a_{n}z^{n} converges uniformly on an open set containing z, which is what we mean by saying that Σ a_{n}z^{n} converges locally uniformly on A.

Since Σ a_{n}z^{n} is a locally uniform limit of continuous functions it is itself continuous.

**Proposition**:

Let f(z) = Σ a_{n}z^{n} on the disc A. Then f is complex differentiable on A, and the derivative is the term-wise derivative Df(z) = Σ na_{n}z^{n-1}.

**Proof**:

First of all we note that Df has the same radius of convergence as f, and so is defined and continuous on A. We can do this either by more or less repeating the proof above, or we can use Hadamard's formula for the radius of convergence. We also note that applying this to Df gives that D^{2}f(z) too is well-defined and continuous on A.

Now for z ∈ A and h sufficiently small

|f(z+h) - f(z) - hDf(z)| =
|Σ a_{n}(z+h)^{n} - a_{n}z^{n} - hna_{n}z^{n-1}| ≤
Σ |a_{n}(z+h)^{n} - a_{n}z^{n} - hna_{n}z^{n-1}| <
Σ h^{2}n(n-1)(z+h)^{n-2} =
h^{2}D^{2}f(z+h)

where we have used the binomial theorem together with the estimate _{n+2}C_{k+2} < n(n-1)_{n}C_{k} in the last inequality. Since D^{2}f is continuous at z we therefore get

|(f(z+h) - f(z))/h - Df(z)| → 0 as h → 0

which implies that f'(z) exists and equals Df(z).

**Corollary**:

A power series f(z) is infinitely differentiable inside its radius of convergence, and f^{(n)}(z) = D^{n}f(z).

**Proof**:

Since D^{n}f(z) is itself a power series we can use induction.