I need a hint to solve this little problem:
Let $f_n:]0,1] \to \Bbb R$ defined by $f_n(x)=x^n (\ln x)^2$. Prove that $$\sum_{n=1}^{\infty} f_n(x)$$ converges uniformly in $]0,1]$.
I started proving uniform convergence in every compact subset $[a,1]$ for $a>0$. (just in case).
Any help? Thank you!
$$f_n'(x) = nx^{n-1}(\log x)^2 + x^n 2 \log(x) \frac{1}{x} = x^{n-1}\log(x)(n \log(x) + 2)$$ So the max on $[0,1]$ (take limits where definitions are excluded) occurs at either $0,1$ or $e^{-2/n} \in (0,1)$. At $0,1$ we have $f_n(x)=0$ so try $f_n(e^{-2/n}) = \frac{4e^{-2}}{n^2}$. Thus $f_n(x) \leq C \frac{1}{n^2}$ so we have a uniform bound by a convergent series.