How find the approximate $\ln{2}$ such the error is less than $0.001$ if $$1.4142<\sqrt{2}<1.4143$$, use it to approximate
$$\ln{2}$$ such the error  is less than $0.001$
This is  National Higher Education Entrance Examination.

Let us try : $$\log(2)=2\log(\sqrt 2)$$ Now, let us use $$\log\frac{1+x}{1-x}=2\Big(\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+...\Big)$$ and use $$\frac{1+x}{1-x}=\sqrt 2$$ that is to say $x=3 - 2 \sqrt 2$, which means, according to $1.4142<\sqrt{2}<1.4143$, $0.1714<x<0.1716$ (this is a small number which will make the series expansion converging quite fast).
If we use the first term only, the result is between $0.3428$ and $0.3432$ (this is already good); using the first two terms, the result is between $0.346157$ and $0.346569$; using the first three terms, the result is between $0.346216$ and $0.346628$.
