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This is a very simple question and still I have a problem with that. I want to check whether this series converges $\sum\limits_{n=1}^{\infty}\frac{x}{1+n^4x^2}$ where $x \in [0, \infty]$

This is an attempt to use the M-Weierstrass test for uniform convergence:

$\sum_{n=1}^{\infty}|\frac{x}{1+n^4x^2} |\leq \sum_{n=1}^{\infty}|\frac{x}{n^4x^2}| \leq \sum_{n=1}^{\infty}|\frac{1}{n^4x}|\leq \frac{1}{|x|} \sum_{n=1}^{\infty}|\frac{1}{n^4}|$ when $x$ is not 0, bus still fixed. For 0, It's easy to see that there's a uniform convergence. I feel that something is wrong with this use of this test. What is wrong? and What is the correct way showing the uniform convergence?

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  • $\begingroup$ I think that's correct for $|x|\geq1$ $\endgroup$ – fiorerb Feb 10 '12 at 15:11
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Since, for every $x$ in $\mathbb R$ and integer $n\geqslant1$, $$ \left|\frac{x}{1+n^4x^2}\right|\leqslant\frac1{2n^2}, $$ the series is normally convergent on $\mathbb R$. In particular, it is uniformly convergent on $\mathbb R$.

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  • $\begingroup$ I feel dumb asking, but why is this inequality holds? $\endgroup$ – Elimination Jul 2 '15 at 16:02
  • $\begingroup$ @Elimination What did you try to prove it? $\endgroup$ – Did Jul 3 '15 at 10:38
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I believe that Dini's theorem is applicable here.

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  • $\begingroup$ Should I need to show that just $a_n$ is a monotic function for $\sum_{n=1}^{\infty} a_n$? $\endgroup$ – Jozef Feb 10 '12 at 15:39
  • $\begingroup$ $[0,\infty)$ is not compact. $\endgroup$ – Najib Idrissi Feb 10 '12 at 16:23
  • $\begingroup$ @zulon The OP has a solution away from zero. The issue is on [0,1]. $\endgroup$ – Mark McClure Feb 10 '12 at 17:11

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