Show $$p< -\log(1-p)$$ for $0<p<1$. I thought about using Taylor expansion, is that the way to go?
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4Possible duplicate of prove $\log(1+x)<x$ for $>0$ – Martin R May 06 '19 at 11:21
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1Or better this one: Show that $\ln (x) \leq x-1 $ – Martin R May 06 '19 at 11:23
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Also closely related (and in fact equivalent): Simplest or nicest proof that $1+x \le e^x$ – Martin R May 06 '19 at 11:26
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You want to prove $f(p):=-p-\ln(1-p)$ is positive on $(0,\,1)$. This follows from $f(0)=0$ and $f^\prime(p)=\frac{p}{1-p}>0$. The Taylor series $f(p)=\sum_{n\ge2}\frac{p^n}{n}$ also works, as you suggested.
J.G.
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You have for $0<p<1$
$$-\log(1-p) = \int_{1-p}^1\frac{dt}{t}> \int_{1-p}^11\;dt = 1-(1-p) = p$$
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