Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 2G / b / ii / Solution

For $0<x<1$, $|x^k/k^2|\le 1/k^2$, and the numerical series ${\sum }_{k\ge 1}{k}^{-2}$ converges. The Weierstrass M-test therefore gives uniform convergence to $f(x)={\sum }_{k\ge 1}{x}^k/k^2$. More explicitly,The bound is independent of $x$, including points arbitrarily close to one.