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$\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{1}{1+n^2x}$ 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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    $\begingroup$ 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. $\endgroup$
    – user29743
    Apr 25, 2012 at 7:53
  • $\begingroup$ Extending Gingerjin's (and Henry's) hint: If $x>1/K>0$, then $0<(1+n^2x)^{-1}<Kn^{-2}$. $\endgroup$
    – Jyrki Lahtonen
    Apr 25, 2012 at 11:33
  • $\begingroup$ ... 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). $\endgroup$
    – Jyrki Lahtonen
    Apr 25, 2012 at 11:36
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    $\begingroup$ Have you covered something called "Weierstrass' M-test" in class? $\endgroup$
    – Jyrki Lahtonen
    Apr 26, 2012 at 12:33
  • $\begingroup$ yes I know that M-test $\endgroup$
    – Marso
    Apr 26, 2012 at 16:20

1 Answer 1

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

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

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