Proving Polynomial is Analytic If a function $f$  at $x = a$ equals it's Taylor Series, $f$ is said to be analytic. 
So, if I were given a polynomial $p(x) = \sum_{n=0}^{200}{a_nx^n}$, and trying to prove that $p(x)$ was analytic at every a $\epsilon$ $R$, could I do something similar to showing that a limit exists:
$\exists$ $\delta$ > 0 s.t $\sum_{n=0}^{200}{a_nx^n}$ converges to $f(x)$ $∀$ $x$ $\epsilon$ ( a - $\delta$, a + $\delta$)?
Or am I misinterpreting what I'm trying to prove?
 A: Actually you should be taking the series at an arbitrary $a$.  To get the series, you write your polynomial in terms of powers of $x-a$ rather than powers of $x$.  All but finitely many terms of the series will be zero, so there's not much problem proving convergence.
A: What you say is lacking a little rigor.
"If a function $f$  at $x = a$ equals it's Taylor Series, $f$ is said to be analytic": actually the equality must hold for a whole interval of values of $x$. The Taylor series at $a$ trivially reduces to $f(a)$ as all other terms are zero. Even a matching on a finite number of points is insufficient, as it could occur "by accident". Functions have an infinity of degrees of freedom, an infinite Taylor development, and matching must be checked on infinitely many points.
"$\exists$ $\delta$ > 0 s.t $\sum_{n=0}^{200}{a_nx^n}$ converges to $f(x)$ $∀$ $x$ $\epsilon$ ( a - $\delta$, a + $\delta$)": this is better, you now have a comparison interval $(a-\delta,a+\delta)$. But it's the Taylor development of the function $f$ that must converge to $f(x)$; in your case $\sum_{n=0}^{200}{a_nx^n}$ is $f(x)$ (!)
Now the answer: a polynomial is certainly an analytic function everywhere, as its Taylor development is formally identical to itself. It is finite so cannot diverge.
Examples:
$p(x)=x^3+x-2$ around $x=0$: 
$$p(0)+p'(0)x+p''(0)x^2/2+p'''(0)x^3/3!+\dots=-2+1.x+0.x^2/2+6.x^3/3!$$
(all higher order terms are $0$.)
$p(x)=x^3+x-2$ around $x=1$: 
$$p(1)+p'(1)x+p''(1)x^2/2+p'''(1)x^3/3!+\dots=0+4.x+6.x^2/2+6.x^3/3!$$
As you can check, $$(x+1)^3+(x+1)-2=x^3+3x^2+4x.$$
