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Posted (edited)

1. The problem statement, all variables and given/known data

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

 

3. The attempt at a solution

 

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

 

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

 

[math]\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. [/math]

 

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

 

[math] 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[/math]

[math]\Rightarrow \mbox { is not uniformly convergent } \in R[/math]

 

What do you think?

[i left out technical details to present my idea more clearly.]

Edited by estro

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