I am really sorry if this question sounds stupid or too obvious. Even though I know how to apply all of these when calculating limits or sums of series, I still have questions that I need to answer.
1) Taylor's theorem says that a k-times differentiable function can be approximated by a k-th order Taylor's polynomial in the neighborhood of some given point.
Okay, this seems absolutely clear. We have a function and we want to linearize this function near some $a$.
2) Taylor's series is a representation of a function that is infinitely differentiable at a real or complex number a.
Should we say this representation is true in the neighborhood of $a$?
Let's say we want to express $e^x$ as a power series (at $a=0$). Do we need to say that this power series holds in the neighborhood of zero? Or just at $0$?
3) If I understand correctly, a power series is technically the same as a Taylor series, but most of the time we are interested in using Taylor's series. Is that correct?
Can you please clarify the questions above for me?
Given any sequence $n\mapsto a_n\in{\mathbb C}$ $(n\geq0)$ of complex numbers you can put on the blackboard the formal power series $$\sum_{n=0}^\infty a_n z^n\ .\tag{1}$$ Such an object makes sense, e.g., in combinatorics, even if this series does not converge for any $z\ne0$. If there is a $\rho>0$ such that $(1)$ converges for $|z|<\rho$ then $$f(z):=\sum_{k=0}^\infty a_n z^n\qquad(|z|<\rho)$$ is a well defined function on the open disc of radius $\rho$. If you are just interested in the real environment and all $a_n\in{\mathbb R}$ then the real-valued function $f$ is well defined on the interval $\>]{-\rho},\rho[\>$. It is then proven in Calculus 102 that this $f$ is in fact infinitely differentiable, and that $$a_n={f^{(n)}(0)\over n!}\qquad(n\geq0)\ .$$ This is saying that the given series is in fact the Taylor series at $0$ of the function it defines: $$f(x)=\sum_{n=0}^\infty{f^{(n)}(0)\over n!} x^n\ .$$ Conversely things are a little different: Given a function $f$ which is $N\geq0$ times differentiable at $0$ we can set up its Taylor polynomial $j_0^Nf$ of degree $N$ at $0$: $$j_0^N f(x):=\sum_{n=0}^N{f^{(n)}(0)\over n!} x^n\ .$$ In Calculus 102 it is shown that this is the unique polynomial $p_N$ of degree $\leq N$ satisfying $$f(x)-p_N(x)=o(x^N)\qquad(x\to0)\ .\tag{1}$$ Since these $j_0^Nf$ approximate $f$ ever better in the neighborhood of $x=0$ when $N\to\infty$ we are led to the question whether in fact $f$ is "represented by its Taylor series", i.e., whether $$f(x)=j_0^\infty f(x):=\sum_{n=0}^\infty{f^{(n)}(0)\over n!} x^n\tag{2}$$ holds in a neighborhood $|x|<\rho$ of $x=0$. In order to answer this question one has to analyze the remainder term $o(x^N)$ in $(1)$ for the particular $f$ under consideration.
An example: If $f(x):=\log(1+x)$ then one needs "Taylor's theorem with remainder" and estimates for $f^{(n)}(x)$ $\bigl(|x|\leq{1\over2}\bigr)$ to prove that $(2)$ is true for this $f$ and $|x|\leq{1\over2}$. One then obtains that indeed $$\log(1+x)=x-{x^2\over2}+{x^3\over3}-\ldots\qquad\bigl(|x|<{1\over2}\bigr)\ .$$ But there are examples of $C^\infty$ functions for which $(2)$ does not hold, e.g., $f(x):=e^{-1/x}$ $(x>0)$ and $:=0$ $(x\leq0)$. I won't go into this.