Question: 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...

Answer: 
Suppose that a sequence $f\_n(x)$ converges pointwise to the function $f(x)$ for all $x\in S$.

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> **NEGATION OF UNIFORM CONVERGENCE**
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> 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 $|f\_{n\_0}(x)-f(x)|\ge \epsilon$.
> 
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> 

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 $x>1/n$.

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**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.
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.