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If $x=c$, then

$$\begin{aligned} f(x)&=A_a(c-c)^a+A_{a+1}(c-c)^{a+1}+A_{a+2}(c-c)^{a+2}+\cdots\\ &=A_a \cdot 0+A_{a+1}\cdot 0+A_{a+2}\cdot 0+\cdots\\ &=0 \end{aligned}$$

So, $f(x)$ converges (to the constant 0) when $x=c$. (Had we allowed $a=0$, it would be possible for $f(x)$ to converge to a nonzero number $A_0$, because we use the convention $0^0=1$.)

Depending on the sequence $\{A_n\}$, it's possible that $f(x)$ diverges for all $x \neq c$. For example, suppose $A_n=n!$, so $f(x)=\displaystyle\sum_{n=0}^\infty n!(x-c)^n$. If $x \neq c$, then the limit $\displaystyle\lim_{n\to\infty} \left|\frac{(n+1)!(x-c)^{n+1}}{n!(x-c)^n}\right|=\lim_{n\to\infty}(n+1)|x-c|$ is infinity, since $x-c\neq 0$. So, the series diverges.

We've now shown that the series definitely converges at $x=c$, but at any other point, it may fail to converge.