Taylor's Theorem tell us that (given some assumptions) if we write a function $f(x)$ as the sum of a Taylor polynomial and a remainder term, the remainder term has a specific form
$$R_{n,a}(x)=\frac{f^{(n+1)}(t)}{(n+1)!}(x-a)^{n+1}$$
where $t \in (a,x)$.
Let $f(x)=\sin{x}$.
If we let $a=0$ and write out the remainder term and estimate it we can show that the sequence of remainders converges to $0$.
My question is: how do we get from the theory above to showing that in fact $\sin{x}$ equals the associated infinite Maclaurin series $\sum\limits_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$?
The fact that the remainders converge to $0$ seems to mean that if we consider the sequence of Taylor polynomials, the infinite series of their partial sums is at least pointwise convergent to $\sin{x}$.
If we can show that this infinite series is uniformly convergent, then we are saying that
$$\lim\limits_{n\to\infty} P_{n,a}(x)=f(x), \text{ for all } x$$
This seems to be the answer to my question. Is it?
We would then write
$$\lim\limits_{n\to\infty} P_{n,0}(x)=\lim\limits_{n\to\infty} \sin{x}+\lim\limits_{n\to\infty} R_{n,0}(x)=\sin{x}$$
Ie
$$\sum\limits_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}=\sin{x}$$
But how do we show this uniform convergence?
I just noticed that if we don't assume $a=0$, then I am not sure what happens.
$\sin{x}$ doesn't seem to be the infinite series of partial sums of Taylor polynomials at some $a\neq 0$.