<s>[INST] Upper bound on $\ln(\frac{1}{1-x})$ for $0\leq x\leq 1/2$ Prove that $$\ln\left(\frac{1}{1-x}\right)\leq x+2x^2$$ for $0\leq x\leq 1/2$.
I thought about the Taylor series $\ln(1+x)=x-x^2/2+x^3/3-\ldots$. For small $x$, the values $1+x$ and $1/(1-x)$ are very close to each other, so the inequality should hold since in the Taylor expansion we have $-x^2/2$ while in the desired inequality we have $2x^2$. However, we need to prove the inequality up to $x\leq 1/2$, so something more is needed.
 [/INST] There are many ways. You suggested series, so let's do it that way.
We have 
$$\ln\left(\frac{1}{1-x}\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots.$$
The part from $\frac{x^2}{2}$ on is $\le \frac{x^2}{2}\left(1+x+x^2+x^3+\cdots\right)$. Note that $1+x+x^2+\cdots=\frac{1}{1-x}$ and $\frac{1}{1-x}\le 2$ on our interval.  That gives an inequality sharper than the proposed one. 
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