Let $\{a_n\}$ be a sequence of real numbers and let $f_n(x) = \sum_{i=0}^{n} a_i x^i$. Suppose $\lim_{n \rightarrow \infty} f_n(1)$ exists. Show that $f_n$ converges uniformly on $[-\frac {1} {2} , \frac {1} {2}]$.
My attempt $:$
Since $\lim_{n \rightarrow \infty} f_n(1)$ exists so the infinite series $\sum_{i=0}^{\infty} a_i$ is convergent and hence $\lim_{n \rightarrow \infty} a_n = 0$. Hence for a given $\epsilon>0$ there exists $k \in \mathbb N$ such that for all $n \ge k$ we have $|a_n|< \frac {\epsilon} {2}$. Now for all $ x \in [-\frac {1} {2}, \frac {1} {2}]$, for all $p \in \mathbb N$ and for all $n \ge k$ we have $|f_{n+p}(x)-f_n(x)|=|\sum_{i=n+1}^{n+p} a_i x^i| \le \sum_{i=n+1}^{n+p} |a_i x^i| \le \sum_{i=n+1}^{n+p} \frac {|a_i|} {2^i} < \frac {\epsilon} {2} \sum_{i=n+1}^{n+p} \frac {1} {2^i}< \frac {\epsilon} {2} \sum_{i=0}^{\infty} \frac {1} {2^i}= \frac {\epsilon} {2} . 2 = \epsilon$.
Hence by Cauchy criterion for uniform convergence we have the given sequnce of functions $\{f_n\}$ converges uniformly on $[-\frac {1} {2}, \frac {1} {2}]$.
Is the above reasoning correct at all? Please verify it.
Thank you in advance.
This is a very easy routine problem.
In fact, you don't need $\lim_{n \to \infty}f_{n}(1)$ exists.
We know that \[ \sum_{n \geq 0} x^{n} = 1/(1 - x) \] converges uniformly in $[-1/2, 1/2]$ and so does its derivative.
Thus, for the uniform convergence over the given interval, coefficients can even be unbounded as n goes to infinity.