# Uniform Convergence of Series

• Nov 3rd 2007, 06:30 PM
tbyou87
Uniform Convergence of Series
Hi,

Show that the series $\sum \frac{x^n}{1+x^n}$ converges uniformly on [0,a] for any a in (0,1)

There is a theorem in the book that says that is the series of functions satisfies the cauchy criterion uniformly on a set S, then the series converges uniformly on S. So I guess i'm supposed to use that. I'm not sure how to do a cauchy criterion proof though.

Thanks
• Nov 3rd 2007, 06:48 PM
kalagota
Quote:

Originally Posted by tbyou87
Hi,

Show that the series $\sum \frac{x^n}{1+x^n}$ converges uniformly on [0,a] for any a in (0,1)

There is a theorem in the book that says that is the series of functions satisfies the cauchy criterion uniformly on a set S, then the series converges uniformly on S. So I guess i'm supposed to use that. I'm not sure how to do a cauchy criterion proof though.

Thanks

we haven't discussed this yet (it would be this coming semester) but im reading in advance.. would Weierstrass M-Test help?
thanks..
• Nov 3rd 2007, 06:56 PM
Jhevon
Quote:

Originally Posted by tbyou87
Hi,

Show that the series $\sum \frac{x^n}{1+x^n}$ converges uniformly on [0,a] for any a in (0,1)

There is a theorem in the book that says that is the series of functions satisfies the cauchy criterion uniformly on a set S, then the series converges uniformly on S. So I guess i'm supposed to use that. I'm not sure how to do a cauchy criterion proof though.

Thanks

Quote:

Originally Posted by kalagota
we haven't discussed this yet (it would be this coming semester) but im reading in advance.. would Weierstrass M-Test help?
thanks..

i was thinking about the Weierstraβ M-Test, but i cannot think of a nice upper bound.

use the theorem that says: If the series $\sum g_n$ converges uniformly on a set S, then $\lim_{n \to \infty} \left[ \sup \{ |g_n(x)| : x \in S \}\right] = 0$

and use the contrapositive

TPH should be able to come up with a more elegant solution though, i'm sure he can use the Cauchy criterion proof/disproof
• Nov 3rd 2007, 07:16 PM
tbyou87
Doesn't that theorem only apply to sequences and not series?
• Nov 3rd 2007, 07:19 PM
Jhevon
Quote:

Originally Posted by tbyou87
Doesn't that theorem only apply to sequences and not series?

it is somewhat different from the series one, and as you can see, the theorem is talking about a series. it turns out that the series wont uniformly converge if the sequence of functions it is summing does not. the theorem was proven in your book as an example, look at it
• Nov 3rd 2007, 07:24 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
Weierstraβ M-Test

:D You started doing German "S" too?

Quote:

TPH should be able to come up with a more elegant solution though, i'm sure he can use the Cauchy criterion proof/disproof
$0\leq \frac{x^n}{x^n+1} \leq \frac{x^n}{1} = x^n \leq a^n$.
But $\sum_{n=0}^{\infty}a^n < \infty$ since $a\in (0,1)$.

Note: Uniform convergence might not work on the full interval $[0,1]$.
• Nov 3rd 2007, 07:30 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
:D You started doing German "S" too?

hehe, yes, and it is for "ss" :D

Quote:

$0\leq \frac{x^n}{x^n+1} \leq \frac{x^n}{1} = x^n \leq a^n$.
But $\sum_{n=0}^{\infty}a^n < \infty$ since $a\in (0,1)$.

Note: Uniform convergence might not work on the full interval $[0,1]$.
how would you show that it's not, without the limit theorem that i posted earlier?
• Nov 3rd 2007, 07:35 PM
kalagota
Quote:

Originally Posted by Jhevon
i was thinking about the Weierstraβ M-Test, but i cannot think of a nice upper bound.

i was thinking if 1 can be an upper bound.. oh well, $a^n$ is.. Ü
• Nov 3rd 2007, 07:38 PM
Jhevon
Quote:

Originally Posted by kalagota
i was thinking if 1 can be an upper bound.. oh well, $a^n$ is.. Ü

the series $\sum_{n = 0}^{\infty} 1$ does not converge, so using that makes no sense. TPH addressed this already, so no worries
• Nov 3rd 2007, 07:48 PM
tbyou87
There is a follow up question that is similar.

Does the series converge uniformly on [0,1)? Prove your idea.

Wouldn't that just be the same proof that TPH did using the weierstrass M-test or am I missing something.

Thanks
• Nov 3rd 2007, 07:52 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
hehe, yes, and it is for "ss" :D

how would you show that it's not, without the limit theorem that i posted earlier?

First I wanted to say $[0,1)$ not $[0,1]$ because it does not converge at $1$. But it does converge (in fact uniformly as shown) on $[0,1)$.

1)The functions $f_n(x) = \frac{x^n}{x^n+1}$ are uniformly continous functions on $[0,1)$.

2)Therefore the series of functions $g(x) = \sum_{n=0}^{\infty}f_n(x)$ is uniformly continous on $[0,1)$.*

3)Thus, if $x_n$ is Cauchy in $[0,1)$ then $g(x_n)$ is also Cauchy (properties of uniformly continous functions). But as you can see as you get closer to $1$ the value of $g(x)$ becomes unbounded. So it cannot be Cauchy.

4)Thus, our initial assumption that the convergence was uniform on the full interval $[0,1)$ was wrong.

*)Because uniformly continous functions that converge uniformly will convergence to a uniformly continous functions.
• Nov 3rd 2007, 08:09 PM
tbyou87
Quote:

Originally Posted by ThePerfectHacker
First I wanted to say $[0,1)$ not $[0,1]$ because it does not converge at $1$. But it does converge (in fact uniformly as shown) on $[0,1)$.

4)Thus, our initial assumption that the convergence was uniform on the full interval $[0,1)$ was wrong.

Sorry I'm a little bit confused. Does the series converge uniformly on [0,1) or not?

Thanks again
• Nov 3rd 2007, 08:11 PM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
Sorry I'm a little bit confused. Does the series converge uniformly on [0,1) or not?

No. (That was not part of your original problem, that was something I added.)