<s>[INST] Bounds for $\log(1-x)$ I would like to show the following

$$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$

I know that for $|x|<1$, we have $\log(1-x)=-\left(x+\frac{x^2}{2}+\cdots\right)$. The inequality on the right follows because the difference is $\frac{x^2}{2}+ \frac{x^3}{3} + \cdots \ge 0$.
For the inequality on the left, the difference is $\frac{x^2}{2}-\left(\frac{x^3}{3}+\frac{x^4}{4} + \cdots\right)$. How do I show this is nonnegative?
 [/INST] $\begin{array}\\
-\ln(1-x)
&=\sum_{k=1}^{\infty} \frac{x^k}{k}\\
&=x+\sum_{k=2}^{\infty} \frac{x^k}{k}\\
&=x+x^2\sum_{k=2}^{\infty} \frac{x^{k-2}}{k}\\
&=x+x^2\sum_{k=0}^{\infty} \frac{x^{k}}{k+2}\\
&\le x+x^2\sum_{k=0}^{\infty} \frac{x^{k}}{2}\\
&\le x+\frac{x^2}{2}\sum_{k=0}^{\infty} x^{k}\\
&= x+\frac{x^2}{2}\frac1{1-x}\\
&= x+\frac{x^2}{2(1-x)}\\
&\le x+x^2 \quad \text{if $0 \le x \le \frac12$}\\
\end{array}
$
</s>