Uniform Convergence of Power Series Let $\sum_{n=0}^\infty a_n \left( x-x_0 \right)^n$ be a power series that converges uniformly over all $x\in \mathbb R$. Prove there exists $N\in\mathbb N$ such that for all $n>N,\; a_n = 0$.
I fail to see how this is true.
From the radius of convergence formula 
$$\frac{1}{R}=0=\limsup_{n\rightarrow\infty}\sqrt[n]{\left|a_n\right|}$$
Now for all $n$ choose $a_n=\frac{1}{n^n}\neq0$ and we get $a_n>a_{n+1}$ so that
$$\limsup_{n\rightarrow\infty}\sqrt[n]{\left|a_n\right|}=\lim_{n\rightarrow\infty}\frac{1}{n}=0$$ We therefore got a convergent series over all the reals, thus uniformly convergent in every $[a,b]\in \mathbb R$ contradicting the above statement...
 A: Suppose that a sequence $f_n(x)$ converges pointwise to the function $f(x)$ for all $x\in S$.

NEGATION OF UNIFORM CONVERGENCE
The sequence $f_n(x)$ fails to converge uniformly to $f(x)$ for $x\in S$ if there exists a number $\epsilon>0$ such that for all $N$, there exists an $n_0>N$ and a number $x\in S$ such that $|f_{n_0}(x)-f(x)|\ge \epsilon$.

Now, let $f_n(x)=a_nx^n$ with $\lim_{n\to \infty}f_n(x)=0$ for all $x\in \mathbb{R}$.  Certainly, either $a_n=0$ for all $n$ sufficiently large or for any number $N$ there exists a number $n_0>N$ such that $a_{n_0}\ne 0$.
Suppose that the latter case holds. Now, taking $\epsilon=1$, we find that
$$|a_{n_0}x^{n_0}|\ge \epsilon$$
whenever $|x|\ge |a_{n_0}|^{-1/n_0}$.  And this negates the uniform convergence of $f_n(x)$.
And inasmuch as the sequence $f_n(x)$ fails to uniformly converge to zero, then the series $\sum_{n=0}^\infty f_n(x)$ fails to uniformly converge.
Note for the example for which $a_n=n^n$ we can take $x>1/n$.

NOTE:
If $a_n=0$ for all $n>N+1$, then we have
$$\sum_{n=0}^\infty a_nx^n = \sum_{n=0}^N a_nx^n$$
which is a finite sum and there is no issue regarding convergence.
A: Let $f$ denote this power series and $(P_n)$ denote the sequence of partial sums. Uniform convergence on $\Bbb R$ implies that $(P_n)$ is uniformly Cauchy. I'll use this in two places. First, there must be some $N$ such that for all $n,m \ge N$, 
$$\|P_n - P_m\|_{\infty} < 1$$
In particular, $\|P_n - P_N\|_{\infty} < 1$ for all $n \ge N$, and since non-constant polynomials are unbounded, $P_n - P_N$ must be constant in $x$, say $:=b_n$. Now,
$$|b_n - b_m| \le \|P_n - P_m\|_{\infty} \to 0 \text{ as $n,m \to \infty$}$$
Where again I used that $(P_n)$ is uniformly Cauchy. Hence $(b_n)$ is Cauchy and so convergent, say to $b$. Thus, $f = b + P_N$ is a polynomial.
A: Uniform convergence of the power series of $f$ is equivalent to the fact that the sequence 
$$
f_n(x)=\sum_{k=0}^na_kx^k, \quad n\in\mathbb N,
$$
is uniformly convergent and equivalently uniformly Cauchy in $\mathbb R$, i.e., for every $\varepsilon>0$, there is an $N$, such that,  $$m,n\ge N\quad\Longrightarrow\quad
\sup_{x\in\mathbb R}\lvert\, f_m(x)-f_n(x)\rvert<\varepsilon.\tag{1}$$
Assume that the power series contains infinitely many terms; say $a_{k_n}\ne 0$, $n\in\mathbb N$, let $\varepsilon>0$ and $N$, such that $(1)$ holds. Then, for $k_n\ge N$ we have
$$
\sup_{x\in\mathbb R}\lvert\, f_{k_n}(x)-f_{k_n-1}(x)\rvert=\sup_{x\in\mathbb R}\lvert a_{k_n}x^{k_n}\rvert=\infty.
$$
Contradiction. Hence, the power series contains only finitely many terms.
A: It appears that you're under the impression that "uniformly convergent" means convergent everywhere in $\mathbb R$.
Consider the sequence of functions $x\mapsto x^n$ on the interval $0<x<1$.
For every value of $x$ in the interval you have $x^n\to0$ as $n\to\infty$.  That is what it means to say this sequence of functions converges pointwise on the interval $0<x<1$.
But the uniform distance from $x\mapsto x^n$ to $x\mapsto0$ is $\displaystyle\sup_{0<x<1} |x^n - 0| = 1$, and that does not approach $0$ as $n\to\infty$.  That is what it means to say this sequence of functions does not converge uniformly on the interval $0<x<1$.
