# Uniform convergence of series

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?

Thanks.

• 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 ... – David Mitra Jun 7 '13 at 20:07
• Oh, you're right. Thank you very much. – user73564 Jun 7 '13 at 21:01

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.
• The value in line three of the sum in line two is off. It should be $x^2{ \color{maroon}{ [1/(1+x^2)^1]} - [1/(1+x^2)^{n+1}\over {1-[1/(1+x^2)] } }$. – David Mitra Jun 8 '13 at 9:33