Let $f_n:[0,1] \rightarrow \mathbb{R}$, $n \geq 1$ be defined as:
$f_n(x) = \cases{ x^n(\log x)^2 & \text{if } x\neq0\cr 0 & \text{if } x=0 }$
prove that $\sum\limits_{n=0}^{\infty}f_n(x)$ converges uniformly in $[0,1]$, and that
$$\int_{0}^{1} \frac{x (\log x)^2}{1-x}=2\sum\limits_{n=2}^{\infty}\frac{1}{n^3}$$
My attempt:
Using l'Hopital I find that $\lim_{x\to 0} x\log x=0$, and obviously $\lim_{x\to 1} x\log x=0$.
Since $f_n$ is continuous for all $n$ and [0,1] is compact $f_n$ is bounded $0 \leq f_n \leq k$ for some $k \in \mathbb{R}$.
And because $f_n < f_m$ in $(0,1)$ and for $n>m$ and $\lim_{n\to\infty}f_n=0$, $f_n$ converges uniformly in $[0,1]$.
Now $f_n(x)=g_n(x)^2=(x^{\frac{n}{2}}\log x)^2$, since we know that $g_n$ converges uniformly to $0$, if I prove that $\sum g_n$ is bounded then $\sum f_n$ converges uniformly by Dirichlet.
So how can I prove that $\sum g_n$ is bounded and how can I prove:
$$\int_{0}^{1} \frac{x (\log x)^2}{1-x}=2\sum\limits_{n=2}^{\infty}\frac{1}{n^3}$$
We can see by summing a geometric series that $\sum f_n$ is point-wise convergent to the function $f$ defined by
$$f(x)=\frac{\ln^2 x}{1-x},\; \forall x\in(0,1)\quad ; \quad f(0)=f(1)=0$$
Moreover we have
$$f'_n(x)=x^{n-1}\log x(n\ln x+2)=0\iff x=e^{-2/n}, x=0,x=1$$ so we see that
$$||f_n||_\infty=f_n(e^{-2/n})=\frac{4e^{-2}}{n^2}$$ hence by the Weierstrass M-test the given series is uniformly convergent on $[0,1]$.