Problem about uniform convergence of series of functions

Question

Prove the following series are not uniformly convergent in $[0,1]$:
\begin{align*}
&1.\quad\sum\limits_{n=0}^\infty x^n\log x\\
&2.\quad\sum\limits_{n=0}^\infty \frac{x^2}{(1+x^2)^n}
\end{align*}
A common way to prove $\sum f_n(x)$ is not uniformly convergent seems to be finding $n_k$ and $x_k$ such that $f_{n_k}(x_k)\to Const\neq0$, that is to prove $f_n$ does not uniformly converge to 0. But for those two problems, the general terms both uniformly converge to 0. Indeed, let $f_n(x)=x^n\log(x)$, $g_n(x)=\frac{x^2}{(1+x^2)^n}$. Then, $f_n'(x)=x^{n-1}(n\log(x)+1)$, $g_n'(x)=\frac{2x}{1+x^2}((1-n)x^2+1)$ which leads to
\begin{align*}
\sup_{x\in[0,1]}|f_n(x)|&=|f(e^{-1/n})|=\frac{e^{-1}}{n}\to0, \text{ as }n\to\infty\\
\sup_{x\in[0,1]}|g_n(x)|&=|g(\sqrt{\frac{1}{n-1}})|=\frac{\frac{1}{n-1}}{(1+\frac{1}{n-1})^n}\to0, \text{ as }n\to\infty.
\end{align*}
So is there any other efficient way to disprove uniform convergence? Thanks!

Answer 1

For the first,

The series converges for $ x\in(0,1]$. the sum function is defined by $$f(x)=\ln(x)\sum_{n=0}^{+\infty}x^n=\frac{\ln(x)}{1-x} \text{ if } x\ne 1$$ and $$f(1)=0$$ but $$\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}\frac{\ln(1-(1-x))}{1-x}=-1$$ thus, As pointed by @Daniel, $ f $ is not continuous at $ (0,1] $ and the convergence is not uniform at $ (0,1]$.

For the second, observe that $$\frac{x^2}{(1+x^2)^n}=\frac{1}{(1+x^2)^{n-1}}-\frac{1}{(1+x^2)^n}$$