Show that $\sum{\frac{x^n}{(1+x^n)^n}}$ converges uniformly 
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. 
 A: 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.
A: 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.  
