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Show that $$\sum{\frac{x^n}{(1+x^n)^n}}$$ converges uniformly on $[0,1].$

I am sorry but for this exercise I got exactly nothing. It seems to be difficult.

Ted Shifrin
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    So far two deleted erroneous solutions. Yes, it's subtle. The Weierstrass M-test fails, as the maximum of the $n$th term basically looks like $\dfrac1{e(n-1)}$, which is not summable. I don't have the right argument for you, yet, however. – Ted Shifrin May 11 '14 at 02:30
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    This problem is driving me nuts. – abnry May 11 '14 at 04:01
  • @Edwin (new name, eh?): Where did you find this problem? – Ted Shifrin May 11 '14 at 15:00
  • @TedShifrin (Edwin is my real name, Cryme was a nick name :) ). It was an oral examination of ENS contest. –  May 11 '14 at 15:14
  • My only two ideas at the moment: split up the sum with bounds dependent on $x$ (most promising idea to me at the present) or find some clever way to rewrite the sum as an integral which allows for easy manipulations. – abnry May 11 '14 at 17:09
  • Well, I've also thought about using Holder's inequality in some clever way. – abnry May 11 '14 at 17:50
  • École Normale Supérieure, @Edwin? Very cool. Une question bien française, il me semble ... :) – Ted Shifrin May 11 '14 at 21:00
  • @TedShifrin Thanks :). Oui les oraux des ENS sont d'une difficulté redoutable (pour moi)! C'est un plaisir de savoir que vous parlez français. –  May 12 '14 at 10:34
  • This was an oral exam? How were you supposed to answer this immediatly, when no one here has a solution after two days? I just...no... – chubakueno May 13 '14 at 05:05
  • @chubakueno Often the candidate is not supposed to solve the problem completely, the goal of the exam is to assess his reactions to very hard problems. I got a very good grade at my oral exam without succeeding to solve the problem – Ewan Delanoy May 13 '14 at 13:27
  • There should be a tag "oral-ens" :-) – Jean-Claude Arbaut May 13 '14 at 13:31
  • Well, @Edwin, now that we see a solution, we see that the problem was "d'une difficulté redoutable" pour tous!! Even my analyst colleagues were stumped. N'ayez pas honte!! – Ted Shifrin May 13 '14 at 16:06
  • @TedShifrin Clearly..You are right, thanks for the valuable comment! –  May 13 '14 at 16:27

2 Answers2

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This problem has already been asked in the American Mathematical Monthly (number of the problem 10840 in the volume 107, number 7, December 2000, page 950) and a solution can be found in the volume 109, number 4 (April 2002), pages 398-399 of the American Mathematical Monthly.

Jean-Claude Arbaut
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user37238
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EDIT: I did quite a basic mistake in the determination of the nth term... Sorry for the answer, which is quite wrong... I put a stop where it is completely off the mark.

Good question, quite interesting. Sorry for the non-LateX writing, but I did not write math LateX since a very long time.

The uniform convergence can be proven with the Cauchy criterion.

In that case, since all terms are positive, the criterion can be reduced to $\sum_{k=N}^{\infty}\frac{x^k}{(1+x^k)^k}$ and show that $\forall \epsilon$ chosen, there can be a N for which, $\forall x$ between 0 and 1, this Sum is < $\epsilon$.

Basically, we are reduced to study the function $F_{k}(x)=\frac{x}{1+x^k}$... To study the derivative is to study the function $G_{k}(x)=x^k-kx+1$. This is fairly easy to show that: for each k>2 there is a 0 for G between 0 and 1 (we note $a_{k}$).

EDIT: This is wrong from now on. We can show that $a_{k}$ ~ $1/k$ for k big enough.

Each term $F_{k}(a_{k})$ ~ $1/k$ as well, for k big enough. Hence, for N big enough, every $F_{k}(x) < 1/2$.

It is straightforward to then see that the Sum of all the terms can be found < $\epsilon$, $\forall x$... Since $\sum_{k=N}^{\infty}\frac{1}{2^k}$ is bigger than the initial sum and can be found as small as we want.

If someone could format it better in math formulas, that would be great... I will try it in the meanwhile.

Martigan
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