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This came up in a part of the proof.

$-\log(1-x)$ is $x$ and then want to calculate the error of this.

The idea is that taylor series of $-\log(1-x)=x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+...$ We have $|x|<1$.I know how to calculate Taylor expansion, however can't see the justification from saying it is x. Next it says what is the error of this.

Well, it has

$x\leq \int_{1-x}^{1} \dfrac{dt}{t} \leq \dfrac{1}{1-x} x$

However, can't understand how this is true.

This is due to trying to prove that $0 \leq \sum_{p\leq N} ((-log(1- \dfrac{1}{p})-\dfrac{1}{p}) \leq \sum_{p \leq N} \dfrac{1}{p(p-1)}$

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    Please dpn't start the question in the title. Keep the body of the question self-contained.2012-02-04
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    @AsafKaragila okay I edited it.2012-02-04

3 Answers 3