Solution
= Solution
For $0<x<1$, $|x^k/k^2|\leq1/k^2$, and the numerical <series> $\sum_{k\geq1}k^{-2}$ converges. The <Weierstrass M-test> therefore gives \b[uniform convergence] to $f(x)=\sum_{k\geq1}x^k/k^2$. More explicitly,
$$
\sup_{0<x<1}|f(x)-f_n(x)|\leq\sum_{k=n+1}^{\infty}\frac1{k^2}\leq\int_n^\infty\frac{dt}{t^2}=\frac1n\longrightarrow0.
$$
The bound is independent of $x$, including points arbitrarily close to one.