Here is Prob. 4, Chap. 7, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Consider $$ f(x) = \sum_{n=1}^\infty \frac{1}{1+n^2 x }. $$ 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?

My Attempt:

Fix a real number $\delta > 0$.

If $x \in [ \delta, +\infty)$, then we see that $$ 0 < \frac{1}{1+n^2 x} < \frac{1}{n^2 x } \leq \frac{1}{n^2 \delta } = \frac{1}{\delta} \frac{1}{n^2}, $$ and the series $\sum \frac{1}{n^2} $ converges. So, by Theorem 7.10 in Rudin, our series converges uniformly on $[ \delta , +\infty)$.

And, if $$x \in (- \infty, - \delta ]\setminus \left\{ \ -1, -\frac{1}{2^2}, -\frac{1}{3^2}, - \frac{1}{4^2}, \ldots \ \right\},$$ then $x < 0$, and, for all large enough $n$, we also have $1 + n^2 x < 0$, and for those $n$, we have
$$ \left\lvert \frac{1}{1+n^2 x} \right\rvert = \frac{1}{ \left\lvert 1+n^2 x \right\rvert} = \frac{1}{ -1 - n^2 x} < \frac{ 1}{ \frac{n^2 x}{2} - n^2 x} = -\frac{2}{x} \frac{1}{n^2} = \frac{2}{\delta} \frac{1}{n^2}. $$ Here we have used the fact that, as $x < 0$, so as $n$ gets larger and larger, we eventually have $1 + n^2 x < -1$, so $n^2 x < -2$, which implies that $\frac{n^2 x}{2} < -1$ and hence $\frac{n^2 x}{2} - n^2 x < -1 - n^2 x$; also from $1 + n^2 x < -1$, we obtain $1 < -1 - n^2 x$ and hence $0 < 1 < -1 - n^2 x$. And, we have also used the fact that, as $x \leq - \delta$, so $-x \geq \delta > 0$, and hence $$ 0 < -\frac{1}{x} \leq \frac{1}{\delta}.$$

As the series $\sum \frac{1}{n^2}$ converges, so it follows from Theorem 7.10 in Baby Rudin that our series converges uniformly on the following subset of $\mathbb{R}$:
$$ (-\infty, -\delta ] \setminus \left\{ \ -1, -\frac{1}{2^2}, - \frac{1}{3^2}, - \frac{1}{4^2}, \ldots \ \right\}.$$

Is what I have done so far correct? If not, then where have I erred?

What if $x \in (- \delta, \delta)$?

And, what about the (uniform) convergence of this series for complex values of $x$?

  • $\begingroup$ With minor modifications, your argument shows uniform convergence for all complex $x$ with $|x|>\delta$, provided $x\neq -1/n^2$. The union of sets $\{x : |x|>\delta\}$ for all positive $\delta$ covers the entire complex plane, except $x=0$. This is all you need to prove continuity everywhere except at those points where the series doesn't converge, i.e. $x=0$ and $\{x=-1/n^2 : n\in\mathbb{N}\}$. The series won't be bounded, though, because it blows up at e.g. $x=0$. $\endgroup$ – Yly Dec 21 '17 at 3:50

Solution I
for values of $x>0$ we have $$\sum\left|\frac{1}{1 + n^2x}\right|= \left|\frac{1}{x}\right|\sum\left|\frac{1}{\frac1x + n^2}\right| \leq \left|\frac{1}{x}\right|\sum\left|\frac{1}{ n^2}\right|$$ By the comparison test this shows that $f(x)$ converges absolutely. If $x = 0$ then we have $$\sum\left|\frac{1}{1 + n^2x}\right| = \sum\left|1\right|= \infty$$

This series clearly doesn’t converge, absolutely or otherwise. If $x < 0$ things get more complicated: If $x = −1/n^2$ for any $n \in \mathbb{N}$ then the nth term of the series is undefined and therefore $f(x)$ is undefined. If $x \neq −1/n^2$ for any $n \in \mathbb{N}$ then we can use the fact that

For what intervals does the function converge uniformly?

If $E$ is any interval of the form $[a, b]$ with $a > 0$ then we have

$$\sup\left|f_n(x)\right| = f_n(a) = \frac{1}{1 + n^2a}$$

And therefore we have $$\sum\sup\left|f_n(x)\right| = \sum\frac{1}{1 + n^2a} \leq \frac1a\sum\frac1{n^2}$$ The complete Answer you can find here (Page. 124)

Here is a second solution to the problem, which I think is more accurate than the previous Solution II See link

  • $\begingroup$ Take a look at the link in the answer, on the page 124 $\endgroup$ – Darío A. Gutiérrez Nov 27 '17 at 13:56

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