Geometric-series bound for the logarithmic remainder ($-\log (1-u)-u<u^2/[2(1-u)]$)

For $0<u<1$, the power series for the logarithm gives $-\log (1-u)-u={\sum }_{n=2}^{\infty }{u}^n/n<\frac{1}{2}{\sum }_{n=2}^{\infty }{u}^n=u^2/[2(1-u)]$. The inequality is strict because $1/n<1/2$ for every $n\ge 3$.