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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.

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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.)

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  • $\begingroup$ Thank you for the answer. I was stupid, I arrived at this result some time ago but thought that converging to a function that satisfies $ \lim_{x\to 1^-} g(x) = +\infty $ is still convergence. (sic!) $\endgroup$
    – zen-dev
    Commented Apr 28, 2013 at 16:37

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