# Power series

### Basic concepts and principles

As we seen at Cauchy's integral formula When a complex function has derivative at a point, it also has infinite derivatives at such point. Then we say that the function is holomorphic at that point.

So, by Taylor's theorem, given a holomorphic function at a point, we can calculate the Taylor series for f of following way: $$f(z) = f(a) + f'(a) (z-a) + \frac{f''(z)}{2!} (z-a)^{2} + ... + \frac{f^{(n)}(z)}{n!} (z-a)^{n} + ....$$

This leads us to a more general type of series of functions called power series, wich is the study of series of the form: $$\sum _{n=0}^{\infty}a_{n}(z-a)^n$$ Note at first that this kind of series are always convergent for z=a.

The previous series converges in some points and diverges in others, in general, there exists a radius R such that the series converges if |z-a| < R and diverges if |z-a| > R. For those points such that |z-a| = R, the series may or not converge. Extreme cases are considered in which R = 0 when the series only converges for z = a or R = ∞. In general, R is called the radius of convergence of the power series.

We show below some tests of convergence for power series:

### Comparison test

If series $\sum |v_{n}|$ is convergent and $|u_{n}| \le |v_{n}|$, then the series given by $\sum u_{n}$ is absolutely convergent.

On the other hand, if the series $\sum |v_{n}|$ diverges and $|u_{n}| \ge |v_{n}|$, then the series $\sum |u_{n}|$ diverges, but series $\sum u_{n}$ can converge or not.

### Quotient test

If $\lim_{n\to \infty} |\frac{u_{n+1}}{u_{n}}| = L$, then $\sum u_{n}$ converges (absolutly) if L < 1 and diverges if L> 1. If L=1 the test fails.

### Root test

If $\lim_{n\to\infty} \sqrt[n]{|u_{n}|} = L$, then $\sum u_{n}$ converges (absolutly) if L < 1 and diverges if L> 1. If L=1 the test fails.

There are some other tests, integral, of Gauss, but with these for now it will be enough for the examples that we will develop here. The following is a test for uniform convergence:

### Test M of Weierstrass

If $|u_{n}(z)| \le M_{n}$ for all z of a region R and $\sum M_{n}$ is convergent, then the series of functions $\sum u_{n}(z)$ it is convergent throughout the region R.

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