Uniform convergence - Series of functions

[estro]

Homework Statement

[tex]\mbox{Check whether } \sum_{n=0}^\infty \frac {1}{e^{|x-n|}} \mbox{ is uniform convergent where its normaly convergent}[/tex]

The Attempt at a Solution

[tex]\mbox{I choose } \epsilon = 1/2[/tex]

[tex]a_n=\frac {1}{e^{|x-n|}}\ ,\ b_n= \frac {1}{n^2}[/tex]

[tex]\lim_{n\rightarrow\infty} \frac {a_n}{b_n}=0\ \Rightarrow\ \sum_{n=0}^\infty a_n \mbox{ is convergent for all x \in R.}[/tex]

[tex]f_k(x)=\sum_{n=0}^{k} a_n\ ,\ f(x)=\lim_{k\rightarrow\infty} \sum_{n=0}^{k} a_n[/tex]

[tex]x_n=n\Rightarrow\\sup_{x_n \in R} |f_k(x_n)-f(x_n)|\geq\sum_{n=k+1}^\infty \frac {1} {e^{|n-n|}}=1>\epsilon\ \Rightarrow \mbox{ is not uniformly convergent \inR}[/tex]

What do you think?
[I left out technical details to present my idea more clearly.]

 

[kurt.physics]

Your attempt at a solution is incorrect. The given series is not uniformly convergent, since there exist points x for which the series does not converge uniformly. To show this, you should take an x such that the terms of the series oscillate in sign near x. For example, if x is an integer, then the series will have an alternating pattern of positive and negative terms near x. This means that the partial sums of the series will oscillate in sign, so the series does not converge uniformly.