# 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.

• 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 – Marso Apr 26 '12 at 16:20

Taking one question more out of the unanswered questions' mud: $$\forall\,0\neq x\in\mathbb R\,\,,\,\,1+n^2x>n^2x\Longrightarrow \frac{1}{1+n^2x}\leq\frac{1}{x}\frac{1}{n^2}$$
Now use Weierstrass's M-test. Note that for $\,x=0\,$ the series trivially diverges.