Test uniform convergence of the series


My attempt:

Here $$f_n(x)=\frac{x}{(n+x^2)^2}$$

Applying Weierstrass's M-test

$$\vert \frac{x}{(n+x^2)^2}\vert < \frac{x}{n^2}<\frac{1}{n^2} \forall x\in(-1,1)$$

Since we know $\frac{1}{n^2}$ is convergent, the series converges for $x\in(-1,1)$

My question is how can I make sure that this is the only interval where the series is Uniformly convergent? And if it isn't, how do I find other such intervals?

  • $\begingroup$ And soon, you will learn about the beautiful function corresponding to the infinite summation. $\endgroup$ – Claude Leibovici Aug 15 '17 at 13:31
  • $\begingroup$ Which one is that? $\endgroup$ – user467745 Aug 15 '17 at 15:00
  • $\begingroup$ $x \psi ^{(1)}\left(x^2+1\right)$ where appears the first derivative of the digamma function. Don't worry : you will learn about it. Its plot is quite nice. $\endgroup$ – Claude Leibovici Aug 15 '17 at 15:14
  • $\begingroup$ @ClaudeLeibovici: Thank you for introducing me to it :) $\endgroup$ – user467745 Aug 15 '17 at 15:18
  • $\begingroup$ You are very welcome ! Use Wolfram Alpha and just type plot x PolyGamma[1, 1 + x^2] for x=-5 to 5 $\endgroup$ – Claude Leibovici Aug 15 '17 at 15:21

With $|x|=y\sqrt{n}$, we have $$\frac{|x|}{(n+x^2)^2}=\frac1{n^{3/2}}\frac{y}{(1+y^2)^2}\le\frac{C}{n^{3/2}}$$ with some positive constant $C$, since $y/(1+y^2)^2$ is continuous for $y>0$ and $\rightarrow0$ as $y\rightarrow\infty$, so it must be bounded. Thus the series is uniformly convergent on the whole real line.

  • $\begingroup$ Thank you for this simple proof $\endgroup$ – user467745 Aug 15 '17 at 13:18

We have, $$|\sum_{n=1}^{\infty} \frac{x}{(n+x^2)^2}|\le C\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}<\infty $$

which give the uniform convergence of the series by Weierstrass test . (see the prove below)

Indeed we let $f_n(x)=\frac{x}{(n+x^2)^2}$ then $$f'_n(x)=\frac{n-3x^2}{(n+x^2)^3} = 0 \Longleftrightarrow x= \pm a_n =\pm\sqrt{\frac{n}{3}}$$

For $x\in (-\infty, -a_n)\cup (a_n, \infty)$ we have $f_n'\le 0$ then $f_n$ is nonincreasing on each interval. Namely,

$$-f_n(a_n)= f_n(-a_n) \le f_n(x)\le f_n(-\infty) =0 \qquad\text{for every $x\in (-\infty, -a_n)$}$$

$$f_n(a_n) \ge f_n(x)\ge f_n(\infty) =0 \qquad\text{for every $x\in (a_n, \infty)$}$$ in any case we have $$|f_n(x)|\le f_n(a_n) \qquad\text{for every $x\in (-\infty, -a_n)\cup (a_n, \infty)$}$$

and for $x\in (-a_n,a_n)$ we have $f_n'\ge 0$ so $f_n$ is nondecreasing on $(-a_n,a_n)$ i.e

$$-f_n(a_n)= f_n(-a_n) \le f_n(x)\le f_n(a_n)\qquad\text{for every $x\in (-a_n, a_n)$}.$$ Therefore for every $x\in\mathbb{R}$ , $$|f_n(x)|\le f_n(a_n) \approx \frac{1}{n^{3/2}}$$

  • $\begingroup$ Thanks a lot for this solution $\endgroup$ – user467745 Aug 15 '17 at 13:19

Let $M$ be a bounded set in $ \mathbb R$. Then there is $c>0$ such that $|x| \le c$ for all $x \in M$. Hence:

$\vert \frac{x}{(n+x^2)^2}\vert \le \frac{|x|}{n^2}<\frac{c}{n^2} \forall x\in M$.

Therefore the seies is uniformly convergent on $M$.

  • $\begingroup$ This points in a misleading direction, the convergence being uniform on the whole real line. $\endgroup$ – Did Aug 15 '17 at 12:55
  • $\begingroup$ Thank you for this solution $\endgroup$ – user467745 Aug 15 '17 at 13:19

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