# On what interval does it converge absolutely, uniformly,fail to converge uniformly?

$\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{1}{1+n^2x}$ would you tell me for what value of $x$ does the series converge uniformly? On what interval does it fail to converge uniformly and absolutely? Is $f$ continuous when the series converges? Is $f$ bounded?

I just able to show that when $x=-1/n^2$ It has problem. will be pleased for answer.

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I had a nonsense answer which the user Henry fixed to the correct answer - hopefully he will come back and post it as an answer. – user29743 Apr 25 '12 at 7:53
Extending Gingerjin's (and Henry's) hint: If $x>1/K>0$, then $0<(1+n^2x)^{-1}<Kn^{-2}$. – Jyrki Lahtonen Apr 25 '12 at 11:33
... and criticizing the phrasing of the question a bit. A series does not converge uniformly at a single point. It simply converges (possibly absolutely) or diverges. Uniform convergence takes place (or not) on a set (typically an interval, but could be a more general set also). – Jyrki Lahtonen Apr 25 '12 at 11:36
Have you covered something called "Weierstrass' M-test" in class? – Jyrki Lahtonen Apr 26 '12 at 12:33
yes I know that M-test – La Belle Noiseuse Apr 26 '12 at 16:20

Observe that, if $x = 0$, $f(x) = \sum_{n}{1} = \infty$. If $x \neq 0$,

$$\large|\frac{1}{1+n^{2}x}\large| \leq |\frac{1}{n^{2}x}|$$

Hence,

$$\sum_{n}{|\frac{1}{1+n^{2}x}\large|} \leq \sum_{n=1}{|\frac{1}{n^2 x}|} \leq \int_{1}^{\infty}{\frac{1}{y^{2}|x|}dy} < \infty$$

Conclude that the series converges absolutely for any $x \neq 0$.

Also observe that, for $x > 0$,

$$f(x) = \sum_{n}{\frac{1}{1+n^{2}x}} \geq \int_{2}^{\infty}{\frac{1}{1+y^{2}x}dy} = \frac{\tan^{-1}(\infty) - \tan^{-1}(2\sqrt{x})}{\sqrt{x}}$$

Since the RHS is unbounded (take $x$ arbitrarily close to $0$), conclude that $f$ is unbounded.

To prove continuity, for $x > 0$, for every $\epsilon > 0$, observe that there exists $n^{*}$, such that, for every $y > x/2$:

$$|f(y) - \sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}y}}| < \frac{\epsilon}{4}$$

$\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}x}}$ is the sum of $n^{*}$ continuous functions and, thefore, is continuous. Conclude that there exists $\delta > 0$, such that, if $|x'-x| < \delta$, $|\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}x}}-\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}x'}}| < \frac{\epsilon}{2}$

Hence, if $|x-y| < \min(\delta, x/2)$,

$$|f(x)-f(y)| \leq |f(x) -\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}x}} +\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}x}} -\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}y}} + \sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}y}}f(y)| \leq$$

And, by triangular inequality,

$$\leq |f(x)-\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}x}}| + |\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}x}} -\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}y}}| + |\sum_{n=1}^{n^{*}}{\frac{1}{1+n^{2}y}}-f(y)| < \epsilon$$

Conclude that $f$ is continuous for $x > 0$. The same proof works for $x < 0$.

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Taking one question more out of the unanswered questions' mud: $$\forall\,0\neq x\in\mathbb R\,\,,\,\,1+n^2x>n^2x\Longrightarrow \frac{1}{1+nx^2}\leq\frac{1}{x}\frac{1}{n^2}$$

Now use Weierstrass's M-test. Note that for $\,x=0\,$ the series trivially diverges.

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