I saw this theorem in Calculus book:
Let $\alpha = \limsup \limits_{n \to \infty} \sqrt[n]{|a_n|}$, $R = \frac{1}{\alpha}, 0 < R^{'} < R$.
$|x| \leq R^{'}\Longrightarrow \Sigma a_n x^n$ is uniformly convergent.
Please tell me an example in which this statement is not true:
$|x| < R\Longrightarrow \Sigma a_n x^n$ is uniformly convergent.
The power series IS uniformly convergent on $\{x: |x|\leq R'\}$.
In the case $0\ne R \ne \infty,$ and $0<R'<R$ we have $1/R'>1/R=\lim \sup_n|a_n|^{1/n}.$ So for all but finitely many $n$ we have $(1/S)^n>|a_n|^n,$ where $S=(R+R')/2.$ So there exists $n_0$ such that, for all $x$ for which $|x|\leq R' ,$ we have $$n>n_0\implies |a_nx^n|\leq (R'/S)^n.$$ Note that $R'/S$ is independent of $x$ and that $0<R'/S<1.$
$R=\infty$ means $\lim_n |a_n|^{1/n}=0.$ If $R=\infty$ and $\infty>R'>0$ take $n_0$ such that $n>n_0\implies |a_n|^{1/n}<1/(2R').$ Then for all $x$ such that $|x|\leq R'$ we have $$n>n_0\implies |a_nx^n|\leq 2^{-n}.$$
Note that all of the above applies verbatim in $\Bbb C$ as well as in $\Bbb R.$ A similar argument shows that $\sum_na_nx^n$ diverges if $|x|>R.$
The power series $\sum_{n=0}^{\infty}x^n$ converges to $\frac {1}{1-x}$ for $|x|<1$ but does not converge uniformly on $\{x : |x|<1\}:$ Let $x_M=\frac {1}{2^{1/(M+1)}}.$ Then $$\sup_{|x|<1}\left|\left(\sum_{n=0}^Mx^n\right)-\frac {1}{1-x}\right|=$$ $$=\sup_{|x|<1}\frac {|x^{M+1}|}{|1-x|}\geq$$ $$\geq \frac {|(x_M)^{M+1}|}{|1-x_M|} =$$ $$=\frac {1/2}{1-x_M}>1/2 .$$