I badly in yet another uniform convergence problem

[end3r7]

I hate, HATE uniform convergence for series... hate this, really do. I'm trying hard, but our prof gives us very difficult borderline problems.

Now, I am almost 100% sure this is not uniformly convergence at x=1, but since it's a stinking series, it's hard to figure out it's summation

Homework Statement

data[/b]
1) Test the following series for Uniform Convergence on [0,1]
<br /> (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}} <br />

The Attempt at a Solution

I honestly have not a clue where to start. I want to try the following, and I don't know if it's true.

<br /> \frac{1}{2} &lt; \frac{1}{1+x^n}<br />
for x in [0,1)

So <br /> (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{2}} &lt;<br /> (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}} <br />

and for x in [0,1) <br /> (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{2}} = \frac{1}{2} <br />

and we know that and for x = 0 <br /> (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}} = 0 <br />

Thus the limit function is not continuous, so it can't be uniformly convergent.

Sound proof or bad proof?

 

[quasar987]

The idea was good, but you assumed that

\sum_{n=1}^{\infty}x^n = \frac{1}{1-x}

while actually this is true if the index starts at n=0 instead of n=1.

 

[end3r7]

The idea was good, but you assumed that

\sum_{n=1}^{\infty}x^n = \frac{1}{1-x}

while actually this is true if the index starts at n=0 instead of n=1.

The logic should still work, however, correct? Or is there no fix?

 

[end3r7]

Isn't that simply going to come out to x/2 then? So it should still work correct?

(The limit function is not continuous, since it's greater than or equal to 1/2 at 1, and not 0)

 

[quasar987]

Now I can't find a fault. :)