# Completing the solution for Baby Rudin Chapter 7 Exercise 4

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.

• What is the domain of $f$? – Brian M. Scott Feb 16 at 4:43
• @BrianM.Scott It's $\mathbb{R}$. The entire problem is situated in $\mathbb{R}$. – Ricky_Nelson Feb 16 at 4:44
• Look at $f(0)$. – Brian M. Scott Feb 16 at 4:46
• At $f(0) = 1+1+1+\dots$. $f$ is clearly unbounded at $x=0$. Is this argument sufficient to prove that $f$ is unbounded. – Ricky_Nelson Feb 16 at 4:50
• @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. – 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 $$\mathbb R \setminus \{0,-1,-\frac 1 {2^{2}},-\frac 1{ 3^{2}},...\}$$.
• What exactly is $k$? Is it just a real number? Did you introduce $k$ because you wanted to work with the partial sum of $f(x)$? – Ricky_Nelson Feb 16 at 5:26
• @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 $\infty$. – Kavi Rama Murthy Feb 16 at 5:28
• Does the first part of your solution imply that $f\left(\frac{1}{k^4}\right) \to \infty$ as $k \to \infty$? – Ricky_Nelson Feb 16 at 6:03