- #1

end3r7

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## Homework Statement

1) Test the following series for Uniform Convergence

[tex]

\sum\limits_{n = 1}^{\inf } {\frac{{( - 1)^n }}{{n^{x}\ln (x)}}}

[/tex]

2) Let f(n,x) = [tex]

\sum\limits_{n = 1}^{\inf } {( - 1)^n (1-x^{2})x^{n}}

[/tex]

a) Test for absolutely convergence on [0,1]

b) Test for uniformly convergence on [0,1]

c) Is [tex]

\sum\limits_{n = 1}^{\inf } {|f(n,x)|}

[/tex] absolutely convergent on [0,1]?

## Homework Equations

## The Attempt at a Solution

For the first, I'm utterly lost. Is there an easy way to deal with such series?

For the second, could I just argue that for all 0<=x<1, there exists a, s.t. x < a <1

and thus

[tex]

|\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}

[/tex]

and for x = 1 and any a > 0

[tex]

|\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}[/tex]

This would prove all 3 right? But can I argue taht way? Can I fix my 'x' ahead of time, or does my argument have to work for all x simultaneously? Cuz if it does, then all I would have to do is choose x between a and 1 and the argument would break down.