O que sria uma expansão por série?

A power series (centered at $a$) is an infinite series of the form $$\sum_{n=0}^\infty c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\cdots$$ Sometimes we'll call this expression a formal power series, since the sum may actually diverge.

We want to define a function $f(x):=\sum_{n=0}^\infty c_n(x-a)^n$, but first we need to figure out the domain, that is, where is $f$ actually defined? In other words, where (i.e. for which $x$) does $\sum_{n=0}^\infty c_n(x-a)^n$ converge?

To analyze convergence, fix a point $x$ and then just treat $\sum_{n=0}^\infty c_n(x-a)^n$ as a series just like before. Since $x$ is fixed, we can treat it like a constant and analyze convergence of the series just like before.

Looking at the $(x-a)^n$ term, which seems pretty geometric, it looks like the Root or Ratio test might be promising. In fact, picking one of these tests is a pretty good place to start. We might be a little concerned about the inconclusive case for these tests, but we can handle those ``edge'' cases with a different test later.

Example 1a: Does $\sum_{n=0}^\infty\frac{x^n}{n!}$ converge for $x=1$?

We can plug in $x=1$ into the sum, so that we're looking at $\sum_{n=0}^\infty\frac{1}{n!}$. This sum converges by either the Ratio or Root test, so yes.

Example 1b. For what $x$ does $\sum_{n=0}^\infty\frac{x^n}{n!}$ converge?

Well, let's apply the Ratio or Root test with $x$ in the series, treating it as just some constant. Here we'll use the Root test: $$\lim_{n\to\infty}\left(\frac{|x^n|}{n!}\right)^{1/n}=\lim_{n\to\infty}\frac{|x|}{n!^{1/n}}=0 <1$$ since $\lim_{n\to\infty}{n!^{1/n}}=\infty$, which can be shown easily with Stirling's formula.

Thus for any fixed $x$, the Root or Ratio test shows that the series always converges, so $\sum_{n=0}^\infty\frac{x^n}{n!}$ converges for all $x\in\mathbb{R}$.

Example 2: For what $x$ does $\sum_{n=0}^\infty\frac{x^n}{n}$ converge?

Let's start by applying the Ratio or root test. $$\lim_{n\to\infty}\left(\frac{|x^n|}{n}\right)^{1/n}=\lim_{n\to\infty}\frac{|x|}{n^{1/n}}=|x|$$ since $\lim_{n\to\infty}n^{1/n}=1$ (use logarithms). You could also apply the Ratio test instead to get the same result. Now this limit is $<1$ iff $|x|<1$, and greater than 1 iff $|x|>1$. So the series converges for $|x|<1$, and diverges for $|x|>1$.

What about for $|x|=1$, i.e. $x=\pm1$? We can't use the Root or Ratio test here since they're inconclusive. Let's look at what the series is when $x=\pm1$. For $x=1$, we get $\sum_{n=0}^\infty\frac{1}{n}$, which is the harmonic series! which we know diverges to $\infty$. For $x=-1$, we get $\sum_{n=0}^\infty\frac{(-1)^n}{n}$, the alternating harmonic series! which we know converges by the Alternating series test. Thus the series converges for $x\in[-1,1)$, and diverges otherwise.

Ready to try some more examples by yourself?

Observation: In the examples, the Root/Ratio test keeps giving us results like "convergence for $|x-a|<R$'', "divergence for $|x-a|>R$'', and "inconclusive for $|x-a|=R$'' (here $a=0$). In the next topic, we'll analyze and formalize this observation further.

A power series centered at $a$ is an infinite series of the form $$\sum_{n\ge0} c_n (x-a)^n = c_0 + c_1 (x-a)^1 + c_2 (x-a)^2 + c_3 (x-a)^3 + \dots,$$ where the terms are functions of $x$, rather than constants. Power series are sort of like polynomials that go on forever, and a polynomial is just a power series that stops at some finite degree! Power series can, on the one hand, be manipulated easily (much like polynomials); on the other hand, they can carry lots of useful information (as we'll see when discussing Taylor series later on). But before getting carried away with these infinite gadgets, let's look at some examples, like the infinite geometric series power series $1+x+x^2+x^3+\dots = (1-x)^{-1}$ centered at $x=0$, which converges at the points $x\in (-1,1)$.