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