Let's assume that $ f(x) = \sum\_{n=1}^\infty a\_n x^n $ has a radius of convergence $1$ 
and that 
$ \lim\_{x\to 1^-} f(x) = +\infty. $


Does it imply that power series $ g(x) = \sum\_{n=1}^\infty (1-x) a\_n x^n $ 
is uniformely convergent 
on $[0,1]$?


Thanks for any help, this one has killed me, I've been trying to solve it for 24 h now. The solution probably exist in some book, but I have been looking for it and neither did I manage to find explicite solution nor did I solve it myself.



Uniform convergence of $g(x) = \sum_{n=1}^\infty (1-x) a_n x^n$
I suspect that putting in a double pole at $x=1$ will show this implication to be false. In other words, take $f(x) = x/(1-x)^2 = \sum\_{n=1}^\infty nx^n$. Then $g(x) = (1-x)f(x)$ still satisfies $\lim\_{x\to1^-} g(x) = \infty$, which should rule out uniform convergence. (Don't know why you're starting your power series at $n=1$, by the way.)


