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For each $n,$ we define \begin{align} f_n:\Bbb{R}\to \Bbb{R} \end{align} \begin{align} x\mapsto f_n(x)=\frac{x^2}{(1+x^2)^n}\end{align} We consider the function \begin{align} f:\Bbb{R}\to \Bbb{R} \end{align} \begin{align} x\mapsto f(x)=\sum^{\infty}_{n=0}f_n(x)=\sum^{\infty}_{n=0}\frac{x^2}{(1+x^2)^n}\end{align}

I want to show that the series does not converge uniformly on $[-1,1]$ but I'm finding it difficult to do that.

First of all, I considered the Weierstrass M test.

MY TRIAL

\begin{align}\left|f_n(x)\right|=\left|\frac{x^2}{(1+x^2)^n}\right|\leq \frac{1}{(1+x^2)^n} ,\;\;\forall \;x\in[-1,1],\;\forall\;n\in\Bbb{N}\end{align}

I'm also thinking that the $\beta_n=\sup\limits_{x\in[-1,1]}|\sum^{n}_{i=0}f_i(x)-\sum^{\infty}_{i=0}f_i(x)|$ approach could be very helpful too!

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    $\begingroup$ Do you have to prove that the sequence $f_n$ converges (this is what you say in the title) or that the series $\sum_{n=0}^\infty f_n$ converges? $\endgroup$ – 5xum Sep 5 '18 at 9:02
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    $\begingroup$ @5xum: The series! $\endgroup$ – Micheal Sep 5 '18 at 9:04
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The convergence is not uniform.

For $x \ne 0$ this is a geometric series:

$$\sum^{\infty}_{n=0}\frac{x^2}{(1+x^2)^n} = x^2\cdot \frac{1}{1-\frac1{1+x^2}}= x^2+1$$

and for $x = 0$ the sum is $0$.

Therefore

$$f(x) = \begin{cases} x^2+1, &\text{ if } x \ne 0\\ 0, &\text{ if }x = 0 \end{cases}$$

Since $f$ is not continuous, the convergence cannot be uniform.

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  • $\begingroup$ Thank you very much but I have shown this before now. I want to show that the series converges uniformly on $[-1,1]$ and not on $\Bbb{R}$ $\endgroup$ – Micheal Sep 5 '18 at 9:18
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    $\begingroup$ @Micheal The convergence is not uniform on $[-1,1]$ so it can't be shown. The sum is the same on $[-1,1]$. $\endgroup$ – mechanodroid Sep 5 '18 at 9:19
  • $\begingroup$ I was thinking so. Is it because of the discontinuity at $0$? $\endgroup$ – Micheal Sep 5 '18 at 9:20
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    $\begingroup$ @Micheal Yes, uniform limit of continuous functions has to be continuous. $\endgroup$ – mechanodroid Sep 5 '18 at 9:21
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    $\begingroup$ If convergence is uniform on $[-1,1]$, sum function $f$ must be continuous! $\endgroup$ – Riemann Sep 5 '18 at 9:21

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