Let $a_n \to 0$. Prove that the series $\sum_{n=0}^\infty a_n x^n$ converges uniformly on $|x|\le 1/2$

Question

Let $\{a_n\}$ be a sequence of real numbers such that $\lim_{n \to \infty} a_n = 0$. Prove that the series $\sum_{n=0}^\infty a_n x^n$ converges uniformly on the closed interval $-1/2 \le x \le 1/2$.

I've just started learning uniform convergence and I am trying to understand what is going on by choosing a specific sequence converging to $0$. So, I've chosen $a_n=1/n$.

How can I show $\sum_{n=0}^\infty \frac{1}{n}\cdot x^n$ converges uniformly?

By the ratio test, I know the series does converge on radius $|x|<1$, but how do I show this converges uniformly?

Answer 1

Such a sequence $\{a_{n}\}$ is bounded, say, $|a_{n}|\leq M$ for all $n$, then M-test gives the result: $|a_{n}||x|^{n}\leq M/2^{n}$ where $\displaystyle\sum_{n}M/2^{n}<\infty$.

Answer 2

In fact the sequence converges normally: there exists $N$ such that $n>N$ implies that $|a_n|<{1\over 2}$.

Let $n>N$ and $|x|\leq {1\over 2}$, We have $|\sum_{i\geq n}a_ix^i|\leq \sum_{i\geq n}{1\over 2}{1\over 2^i} {1\over 2}={1\over 2^{n+2}} {1\over {1-{1\over 2}}} ={1\over 2^{n+1}}$.