# Thread: Proving a function is continuous using uniform convergence

1. ## Proving a function is continuous using uniform convergence

On the domain $\displaystyle [-1,1)$, is $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}x^{n}$ continuous?

If we let $\displaystyle f_{k}(x) = \sum_{n=1}^{k}\frac{1}{n}x^{n}$, then it is clear that for all $\displaystyle k\in \mathbb{N}$, $\displaystyle f_{k}(x)$ is continuous, so now it will suffice to show that $\displaystyle f_{k}(x)$ uniformly converges to $\displaystyle f(x)$ for $\displaystyle x\in [-1,1)$. So for any $\displaystyle \epsilon > 0$, there exists $\displaystyle N$ such that if $\displaystyle n>N$, then $\displaystyle |f_{k}(x) - f(x)| < \epsilon$.

This is where I am running into some trouble because $\displaystyle |f_{k}(x) - f(x)| = \sum_{n=k}^{\infty}\frac{1}{n}x^{n}$. After this point I am not exactly sure where to proceed. I have a hunch that the fact that $\displaystyle f(x)$ converges on that domain will come into play, but I am not sure. Thanks.

2. Originally Posted by Pinkk
On the domain $\displaystyle [-1,1)$, is $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}x^{n}$ continuous?

If we let $\displaystyle f_{k}(x) = \sum_{n=1}^{k}\frac{1}{n}x^{n}$, then it is clear that for all $\displaystyle k\in \mathbb{N}$, $\displaystyle f_{k}(x)$ is continuous, so now it will suffice to show that $\displaystyle f_{k}(x)$ uniformly converges to $\displaystyle f(x)$ for $\displaystyle x\in [-1,1)$. So for any $\displaystyle \epsilon > 0$, there exists $\displaystyle N$ such that if $\displaystyle n>N$, then $\displaystyle |f_{k}(x) - f(x)| < \epsilon$.

This is where I am running into some trouble because $\displaystyle |f_{k}(x) - f(x)| = \sum_{n=k}^{\infty}\frac{1}{n}x^{n}$. After this point I am not exactly sure where to proceed. I have a hunch that the fact that $\displaystyle f(x)$ converges on that domain will come into play, but I am not sure. Thanks.
I assume that you don't know that a power series is uniformly convergent on it's radius of convergence, huh?