How can I find the value of $\ln( |x|)$ without using the calculator? I want to know if there is a way to find for example $\ln(2)$, without using the calculator ?
Thanks 

We can represent the logarithm of positive rational numbers as follows.
First, consider the following null conditionally convergent series (cancelled harmonic series):
$$0=(1-1)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...$$
Note that we are computing $0=\log(1)=\log\left(\frac{1}{1}\right)$ by adding consecutive terms with 1 positive fraction and 1 negative fraction each, taken from the inverses of non-zero integers. This observation may sound trivial now, but it is interesting for what comes next.
We can rearrange the terms of this series to compute $\log(2)$ by taking two positive fractions and one negative for each term.
$$\log\left(2\right)=\left(1+\frac{1}{2}-1\right)+\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{2}\right)+\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{3}\right)+\left(\frac{1}{7}+\frac{1}{8}-\frac{1}{4}\right)+...$$
This can be easily seen to be the Mercator series in disguise, so we have discovered nothing new yet.
But there is more. Similarly, we have
$$\log\left(3\right)=\left(1+\frac{1}{2}+\frac{1}{3}-1\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{2}\right)+\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{3}\right)+\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{4}\right)+...$$
This pattern holds for all positive integers, so the next step is applying the property that $\log(p/q)=\log(p)-\log(q)$ on these representations.
This leads to $\log(p/q)$ by adding $p$ positive fractions and $q$ negative fractions at each step. For example, we have
$$\log\left(\frac{3}{2}\right)=\left(1+\frac{1}{2}+\frac{1}{3}-1-\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{5}-\frac{1}{6}\right)+...$$
as illustrated in http://oeis.org/A166871.
See also Do these series converge to logarithms? and Series for logarithms for further discussion of generalized Mercator series.
