Baby Rudin Chapter 7 Exercise 4

Consider \begin{equation*} f(x) = \sum_{n=1}^\infty \frac{1}{1+n^2x} \end{equation*} For what values of $x$ does the series converge absolutely? On what intervals does it converge uniformly? On what intervals does it fail to converge uniformly? Is $f$ continuous wherever the series converges? Is $f$ bounded?

Can someone please provide some hints on how this problem can be solved? Based on a few sketches for certain values of $n$, I think $f(x)$ is unbounded, but I am having trouble showing that rigorously. Please don't provide complete solutions, I'd much rather fill in the details of the hints. Thanks.

  • $\begingroup$ What is the domain of $f$? $\endgroup$ – Brian M. Scott Feb 16 at 4:43
  • $\begingroup$ @BrianM.Scott It's $\mathbb{R}$. The entire problem is situated in $\mathbb{R}$. $\endgroup$ – Ricky_Nelson Feb 16 at 4:44
  • $\begingroup$ Look at $f(0)$. $\endgroup$ – Brian M. Scott Feb 16 at 4:46
  • $\begingroup$ At $f(0) = 1+1+1+\dots$. $f$ is clearly unbounded at $x=0$. Is this argument sufficient to prove that $f$ is unbounded. $\endgroup$ – Ricky_Nelson Feb 16 at 4:50
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    $\begingroup$ @Randall No domain was specified in the question but Ch. 7 in Rudin deals only with $\mathbb{R}$, so I assumed that the domain is $\mathbb{R}$. I've edited my post to include the full, original question from Rudin. $\endgroup$ – Ricky_Nelson Feb 16 at 5:02

Partial answer: To show that $f$ is unbounded note that $f(\frac 1 {k^{4}}) \geq \sum\limits_{n=1}^{k}\frac {k^{4}} {k^{4}+n^{2}} \geq \sum\limits_{n=1}^{k}\frac {k^{4}} {k^{4}+k^{2}} =\frac {k^{5}} {k^{4}+k^{2}} \to \infty$ as $k \to \infty$.

EDIT The series converges for any $x \in \mathbb R \setminus \{0,-1,-\frac 1 {2^{2}},-\frac 1{ 3^{2}},...\}$. It converges uniformly on the intersection of $\{x: |x|>\epsilon\}$ with $\mathbb R \setminus \{0,-1,-\frac 1 {2^{2}},-\frac 1{ 3^{2}},...\}$ for any positve number $\epsilon$ (by comparison with $\sum \frac 1 {n^{2}}$). This also implies that $f$ is continuous on $\mathbb R \setminus \{0,-1,-\frac 1 {2^{2}},-\frac 1{ 3^{2}},...\}$.

  • $\begingroup$ What exactly is $k$? Is it just a real number? Did you introduce $k$ because you wanted to work with the partial sum of $f(x)$? $\endgroup$ – Ricky_Nelson Feb 16 at 5:26
  • $\begingroup$ @Ricky_Nelson The inequality I wrote is true for any pisitve integer $k$. So I have produced a sequence of numbers in the domain of $f$ such that the values of $f$ at these points tends to $\infty$. $\endgroup$ – Kavi Rama Murthy Feb 16 at 5:28
  • $\begingroup$ Does the first part of your solution imply that $f\left(\frac{1}{k^4}\right) \to \infty$ as $k \to \infty$? $\endgroup$ – Ricky_Nelson Feb 16 at 6:03
  • $\begingroup$ @Ricky_Nelson Yes, exactly. $\endgroup$ – Kavi Rama Murthy Feb 16 at 6:07
  • $\begingroup$ I think I finally understood the complete solution, thanks! $\endgroup$ – Ricky_Nelson Feb 16 at 6:08

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