We consider the sequence $(f_k)$ defined by $f_k:\mathbb{R}\to\mathbb{R}$, $f_k(x)=\dfrac{x^2}{(1+x^2)^k}$ for each $k\in \mathbb{N}$.

I already proved that $\displaystyle\sum_{k=1}^{\infty}f_k$ converges pointwise in $\mathbb{R}$.

My question is, does $\displaystyle\sum_{k=1}^{\infty}f_k$ converge uniformly?


  • 1
    $\begingroup$ You can compute the limit function, $f$, explicitly: $f(0)=0$. For $x\ne0$, using the formula for the sum of a Geometric series, we have $f(x)=1$. Note $f$ is not continuous ... $\endgroup$ – David Mitra Jun 7 '13 at 20:07
  • $\begingroup$ Oh, you're right. Thank you very much. $\endgroup$ – user73564 Jun 7 '13 at 21:01

A full answer to this question (it was already partially answered in the comments):

Define $$ g_n(x)=\sum_{k=1}^{n}f_k(x)\\ =\frac{x^2}{x^2+1} \sum_{k=0}^{n}\frac{1}{(1+x^2)^k}\\ =\frac{x^2}{x^2+1}\frac{1-[1/(1+x^2)]^{n+1}}{1-[1/(1+x^2)]}\\ =\frac{x^2}{x^2+1}\frac{1-[1/(1+x^2)]^{n+1}}{[x^2/(1+x^2)]}\\ =(1-[1/(1+x^2)]^{n+1}) $$

The pointwise limit of this function is $$ g(x)=\begin{cases}1 & x≠0 \\ 0 & x=0 \\ \end{cases} $$ Since this continuous sequence approaches a non-continuous pointwise limit, we may conclude that the sequence does not converge uniformly.

(Thanks for the correction David)

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    $\begingroup$ The value in line three of the sum in line two is off. It should be $x^2{ \color{maroon}{ [1/(1+x^2)^1]} - [1/(1+x^2)^{n+1}\over {1-[1/(1+x^2)] } }$. $\endgroup$ – David Mitra Jun 8 '13 at 9:33
  • $\begingroup$ Good catch; I was wondering why our answers were different. $\endgroup$ – Omnomnomnom Jun 9 '13 at 16:25

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