# Proving a function is continuous using uniform convergence

• Apr 12th 2010, 04:33 PM
Pinkk
Proving a function is continuous using uniform convergence
On the domain $[-1,1)$, is $\sum_{n=1}^{\infty}\frac{1}{n}x^{n}$ continuous?

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

This is where I am running into some trouble because $|f_{k}(x) - f(x)| = \sum_{n=k}^{\infty}\frac{1}{n}x^{n}$. After this point I am not exactly sure where to proceed. I have a hunch that the fact that $f(x)$ converges on that domain will come into play, but I am not sure. Thanks.
• Apr 12th 2010, 09:00 PM
Drexel28
Quote:

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

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

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