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Does the series $\sum \frac{x^n}{1+x^n}$ converge uniformly on $x\in [0,1)$?

I have no idea where to start. Could somebody give me a hint ?

Edit: Could I use something like this ?
$$\left|\frac{x^n}{1+x^n}\right|\leq x^n$$ Because $x<1$, the geometric series $\sum x^n$ converges. Therefore by the M-test we get that the series converges uniformly.

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It does not work. In the M-test, you need to find an upper bound which does not depend on $x$. Here it only gives uniform convergence on $[0,a]$ for any $a<1$. – 1015 Feb 27 '13 at 13:33
@julien Hm... schade... Any better ideas ? ^^ – Kasper Feb 27 '13 at 13:36
@DavidMitra So your line of reasoning is: If the series was uniform convergent, it would have a continuous limit function $f$. But then it would also be continuous in a point very close to $1$, but as this is not true, we get a contradiction, right ? – Kasper Feb 27 '13 at 13:48
@Kasper You don't even need to talk about the limit $f$. Just work on the Cauchy criterion. See my edit. – 1015 Feb 27 '13 at 13:49
Better: The terms $x^n\over 1+x^n$ do not converge to the zero function uniformly on $[0,1)$. Thus the series does not converge uniformly on $[0,1)$. – David Mitra Feb 27 '13 at 13:53
up vote 7 down vote accepted

The partial sums $S_N(x)=\sum_{n=0}^N \frac{x^n}{1+x^n}$ converge pointwise on $[0,1)$ to $S(x)=\sum_{n\geq 0} \frac{x^n}{1+x^n}$ by comparison with the geometric series $\sum x^n$.

Now $$ S(x)-S_N(x)=\sum_{n\geq N+1} \frac{x^n}{1+x^n}\geq \frac{x^{N+1}}{1+x^{N+1}} $$ for all $x\in[0,1)$.

So, letting $x$ tend to $1$, we get $$ \sup_{[0,1)}S-S_N\geq \frac{1}{2} $$ for all $N$.

Hence the convergence to $S$ is not uniform (which by definition is $\sup_{[0,1)}|S-S_N|\longrightarrow 0$ as $N\rightarrow +\infty$.)

Note: you don't even need $S$, you could simply do it with the Cauchy criterion, if you use the fact that $C([0,1),\mathbb{R})$ is complete when equipped with the uniform norm.

Then $$ |S_M(x)-S_N(x)|\geq \frac{x^M}{1+x^M} $$ for all $M>N$ and all $x\in[0,1)$. Hence $$ \sup_{[0,1)}|S_M-S_N|\geq \frac{1}{2} $$ and the sequence is not uniformly Cauchy.

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I don't really understand. What do you mean with $S_M$ and $S_n$ without the $(x)$ ? – Kasper Feb 27 '13 at 14:01
@Kasper $\|f\|_\infty=\sup_{[0,1)}|f|=\sup_{x\in[0,1)}|f(x)|$. – 1015 Feb 27 '13 at 14:03
aah okay. But why must $\sup_{[0,1)}|S_M-S_N|\geq \frac{1}{2}$ go to zero ? – Kasper Feb 27 '13 at 14:04
Okay, some brain malfunction, I understand it now. Thanks for your helping me out :) – Kasper Feb 27 '13 at 14:19
By the way, is the prove of @sbr, also correct ? – Kasper Feb 27 '13 at 14:19

By definition the serie $\displaystyle\sum\frac{x^n}{1+x^n}$ converges uniformly on $[0,1)$ if $$\lim_n\sup_{x\in[0,1)} \sum_{k=n+1}^{\infty}\frac{x^k}{1+x^k}=0.$$

We have $$\sum_{k=n+1}^{\infty}\frac{x^k}{1+x^k}\geq \sum_{k=n+1}^{\infty}\frac{x^k}{2}=\frac{1}{2}\frac{x^{n+1}}{1-x},$$

So it's clear that $\displaystyle\sup_{x\in[0,1)} \frac{x^{n+1}}{1-x}=+\infty, $ and hence the condition of uniform convergence is not verified.

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I don't know of the definition you gave. The cauchy criterion I know of is: The sequence of partial sums of a series $\sum g_k$ of functions is uniformly Cauchy on a set $S$ if and only if the series satisfies the Cauchy criterion: $\forall \epsilon >0 \exists N \forall x\in S:(n\geq m>N \implies |\sum_{k=m}^ng_k(x)|<\epsilon$ – Kasper Feb 27 '13 at 14:07
Oh wait this implies $|g_n(x)|<\epsilon$ for all $x\in S$, which implies $\sup\{|g_n(x)|:x\in S\}\leq \epsilon$ .... aah. right ? – Kasper Feb 27 '13 at 14:14
aaaah. things begin to make sense. So then you $\forall \epsilon >0\exists N\forall n>N :|\sup\{|g_n(x)|:x\in S\}-0|<\epsilon$ Wich proves $\limsup\{|g_n(x)|:x\in S\} = 0$ – Kasper Feb 27 '13 at 14:17
@Kasper The serie converges uniformly if the partial sum $S_n$ converge with the uniform norm to the sum $S$ i.e. $||S_n-S||_{\infty}=\sup_x|S_n(x)-S(x)|\rightarrow 0$. – user63181 Feb 27 '13 at 14:56

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