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Prove or disprove:
If $f(x)=\sum_{k=1}^\infty a_k(x-x_0)^k$ converges on $(a,b)$,then $f(x)$ converges uniformly on $(a,b)$.
I think that $f(x)$ dosen't converge uniformly on $(a,b)$.
Suppose that $f(x)=\sum_{k=0}^\infty x^k$
So $f(x)$ converges on $(-1,1)$ but not uniformly converges on $(-1,1)$
How to explain or disprove it?
It is similar to Abel's Theorem.
Abel's Theorem
If $f(x)=\sum_{k=1}^\infty a_k(x-x_0)^k$ converges on $[a,b]$,then $f(x)$ converges uniformly on $[a,b]$.
Your counterexample works, and here is an explanation. For the series $\sum_{k=0}^\infty x^k$ to converge uniformly on $x\in(-1,1)$, one needs the tail $\sum_{k=N}^\infty x^k$ to be uniformly small on $x\in(-1,1)$ as $N$ goes to infinity. But one sees that as $x$ is very close to 1, the tail can be arbitrarily large no matter how large is $N$, thus it is not uniformly small on $x\in(-1,1)$ as $N\to\infty$.
