Series of a function

Question

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

For what values of $x$ does the series converge absolutely?

For values $x>0$ we have $$\sum\left\vert\,\frac{1}{1+n^2x}\,\right\vert\ =\ \left\vert\,\frac{1}{x}\,\right\vert \sum\left\vert\,\frac{1}{1/x + n^{2}}\,\right\vert\ \leq\ \left\vert\,\frac{1}{x}\,\right\vert \sum\left\vert\,\frac{1}{n^2}\,\right\vert $$

By the comparison test this shows that $f(x)$ converges absolutely. If $x=0$ then we have $$\mid\frac{1}{x}\mid \sum \mid\frac{1}{1/x+n^2}=\sum \mid1\mid= \infty$$

This series clearly doesn't converge, absolutely or otherwise. If $x<0$ let us consider two cases:

$\bullet$ If $x=-\frac{1}{n^2}$ for any $n \in \mathbb{N}$ then the nth term of the series is undefined and therefore $f(x)$ is undefined.

$\bullet$ If $x\neq-\frac{1}{n^2}$: Im having trouble with this case.

On what intervals does it converge uniformly?

I considered the cases where E is any interval of the form $[a,b]$ with a>0 (Didn't have trouble with that). However, I am having trouble when E is any interval of the form $[a,b]$ with b<0..

On what intervals does it fail to converge uniformly? (Did this)

However had trouble with the two last questions: Is $f$ continuous wherever the series converges? Is f bounded ?

Answer 1

I'm going to keep my original post (cus I like the Cauchy test), none the less it suffices to prove $$\int_0^\infty \frac{dn}{1+n^2x}<\infty$$ for fixed $x\in \mathbb{R}^+$.

By the method of Wolfram Alpha

we can show the above is equal to

$$\frac{\pi}{2\sqrt{x}}$$

and this integral "makes sense" as long as their is no division by zero in the integrand which we should know a priori (and you posted in comments above).

Now.... interestingly,

$$\int_{1/10}^\infty \frac{dn}{1+n^2x} = \frac{\tan^{-1}(\frac{1}{10\sqrt{x}})}{\sqrt{x}}<\infty$$

provided $\frac{1}{x} \ge -1/100 \iff -100 \le x$

Hum.... see the trend? If we want to pick any negative $x$ we just need to integrate differently (closer to zero). The only concern is this converges to complex numbers (I'll remove the additions if this type of convergence isn't allowed)

Original post:

Suppose that $x>0$ (I haven't considered the other half yet sorry), then the terms

$$a_n = \frac{1}{1+n^2x}$$

form a positive non-increasing sequence ; hence by the Cauchy condensation test, the series $$f(x) = \sum_{n=1}^\infty \frac{1}{1+n^2x}$$

converges if and only if the series

$$\tilde{f} = \sum_{n=1}^\infty 2^n \frac{1}{1+2^{2n}x}$$ is convergent. But some calculations show us $$\tilde{f} = \sum_{n=1}^\infty \frac{2^n }{2^{2n}(x+ \frac{1}{2^{2n}})} =\sum_{n=1}^\infty \frac{1}{2^{n}(x+ \frac{1}{2^{n}})} \le \sum_{n=1}^\infty \frac{1 }{2^{n}(x+ 1)}= $$ $$=\frac{1}{(x+ 1)}\sum_{n=1}^\infty \frac{1 }{2^{n}}= \frac{1}{x+1}$$

Then by the comparison test $\tilde{f}$ converges (and hence $f$ by Cauchy condensation).

Answer 2

Let $I$ be an interval. If the adherence of $I$ contains a value $t$ equal to $0$ or of the form $-\frac{1}{k^2}$, then obviously $f$ will take arbitrarily large values near $t$, so $f$ cannot be bounded on $I$.

On the other hand, one has a normal convergence (entailing uniform convergence and boundedness) on every interval staying away from those singularities.

To see that, suppose that the adherence of $I$ does not contain any of the singularities, then $I$ must be of the form $[a,b]$ with $a >0$ or $[a,b]$ with $b < -1$.

In the first case, let $n_0$ be an integer larger than $\sqrt{\frac{2}{a}}$. Then for $n\geq n_0$ we have

$$ \big|\frac{1}{1+n^2x}\big| =\frac{1}{1+n^2x} \leq \frac{1}{1+n^2a} \leq \frac{1}{n^2(\frac{a}{2})} =\frac{2}{a} \frac{1}{n^2} $$

This shows that the series converges normally on $I$.

In the second case, let $n_0$ be an integer larger than $\sqrt{\frac{2}{|b|}}$. Then for $n\geq n_0$ we have

$$ \big|\frac{1}{1+n^2x}\big| =\frac{-1}{1+n^2x} \leq \frac{-1}{1+n^2b}=\frac{1}{-1+n^2|b|} \leq \frac{1}{n^2(\frac{|b|}{2})} =\frac{2}{|b|} \frac{1}{n^2} $$

This shows that the series converges normally on $I$.