Is a power series uniformly convergent in its interval of convergence?
No. Think about the geometric series, which converges (but not uniformly) to $$ 1 + x + x^2 + \cdots = \frac{1}{1-x} $$ on $(-1,1)$.
No. Example: $ \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $|x|<1$.
Suppose that $s_n(x):=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}$ converges uniformly on $(-1,1)$ to $\frac{1}{1-x}$.
Then , to $ \epsilon=1$ we get $N \in \mathbb N$ such that
$\frac{|x|^{N+1}}{1-x}=|s_N(x)-\frac{1}{1-x}|<1$ for all $x \in (-1,1)$.
Hence $\lim_{x \to 1-}\frac{|x|^{N+1}}{1-x} \le 1$. But $\lim_{x \to 1-}\frac{|x|^{N+1}}{1-x}= \infty$, a contradiction.