1
$\begingroup$

Prove that the series $\sum_{n=1}^{\infty} \frac{x^n}{n^4}$ defines a continuous function on $[-1 , 1]$.

Lemma I am trying to use: Let $\sum_{n=1}^{\infty}f_n$ be a series which converges uniformly on $A \subseteq \mathbb{R}$ to $f$. If each $f_n$ is continuous on $A$, then $f$ is continuous on $A$.

I am confused on how to prove this. First, we need to show that this series converges uniformly on $[-1, 1]$. So notice that \begin{align*} 0 \leq \left|\frac{x^n}{n^4}\right| = \frac{|x|^n}{n^4} \leq \frac{1}{n^4} \end{align*} for all $x \in [-1,1]$. Then, notice that $\sum_{n=1}^{\infty} \frac{1}{n^4}$ converges since it is a p-series with p = 4. Therefore, by the Weierstrass M-Test, it follows that $\sum_{n=1}^{\infty} \frac{x^n}{n^4}$ converges uniformly on $[-1,1]$. I don't know where to go from here.

1 Answers 1

1

You're almost finished. Inasmuch as $x^n$ is continuous on $A=[-1,1]$ for all $n\in \mathbb{N}$, $f_n(x)=\frac{x^n}{n^4}$ is likewise.

And you have shown the series, $f(x)=\sum_{n=1}^\infty\frac{x^n}{n^4}$, uniformly converges on $A$.

Now apply the Lemma in the OP and you are done!

  • 0
    Ok thanks, that makes sense! – 2017-01-16
  • 0
    You're welcome. My pleasure. -Mark – 2017-01-16
  • 0
    Quick question since you were helpful in the OP. I have to do the same thing as above but with the series $\sum_{n=1}^{\infty} nx^n$ on $(-1 , 1)$. I am having trouble showing this series converges uniformly. Any hints? Thanks. – 2017-01-16
  • 0
    It converges uniformly on any closed interval $[a,b]$ where $-1$|x^n|\le \left(\max(|a|,|b|)\right)^n$ and $\sum_{n=1}^\infty n\left(\max(|a|,|b|)\right)^n$ converges (use the root test or ratio test, for example). – 2017-01-16
  • 0
    Ok, so in previous work that I have done, I have already showed that $\sum_{n=0}^{\infty} x^n$ converges uniformly on any compact subset of $(-1,1)$ which is basically what you said. So do I even need to show anything? – 2017-01-16
  • 0
    Yes. Show that $\sum_{n\ge1}nx^n$ converges on any compact subset of $(-1,1)$. That should be easy. – 2017-01-16