Let $\{a_k\}_{k=1}^\infty \subset \mathbb R$ be a bounded sequence.
Define $\{f_N : (-1,1)\to \mathbb R\}_{N=1}^\infty$ as $f_N(x)=\sum_{k=1}^N a_k x^k$ and define $f:(-1,1)\to \mathbb R$ as $f(x)=\sum_{k=1}^\infty a_k x^k$.
Then, does $\{f_N\}$ converge to $f$ uniformly on $(-1,1)$ ?
I found that for arbitrary $R\in(0,1)$ ; $f_N \to f$ uniformly on $[-R,R]$, but I think this doesn't imply $f_N\to f$ uniformly on $(-1,1)$. (For example, $f_k(x)=x^k$ converges $f(x)=0$ uniformly on arbitrary conpact sets in $(-1,1)$ but $f_k$ doesn't converges to $f$ uniformly on $(-1,1)$.)
And I also have the inequality $$|f_N(x)-f(x)|\leqq \frac{|x|^{N+1}}{1-|x|}M,$$ where $M$ is a bound of $\{a_k\}$. Does this works ?
The convergence need not be uniform. Here's a counterexample:
$$a_n=1\implies f_N(x)=\sum_{n=0}^N x^n\text{, }f(x)=\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$
If the convergence of $\{f_N\}_{N=0}^\infty$ to $f$ was uniform, then for any fixed $\varepsilon>0$,
$$|f_N(x)-f(x)|<\varepsilon\text{ for every }x\in(-1,1)\text{ and every }N\text{ sufficiently large}$$
But taking $\varepsilon=1$, we see that this implies $|f_N(x)-f(x)|<1$ for $N$ sufficiently large, that is, $f_N-f$ is bounded. This is not possible because $\lim_{x\to 1^-}f(x)=\infty$.