Permute order of summation change the sum : $\sum_{n=1}^\infty \frac{(-1)^n}{n}$. In a course of analysis it's written that if $\sum_{k=1}^\infty x_k$ converge but it's not absolutely convergent, then changing any order of summation will change the sum. 
For example, if I consider $$\sum_{n=1}^\infty \frac{(-1)^n}{n}=\ln(2).$$
What will be $-\frac{1}{2}+1+\sum_{n=3}^\infty \frac{(-1)^n}{n}$? To me it should be $$-\frac{1}{2}+1+\ln(2)-1+\frac{1}{2}=\ln(2),$$
so may be it's only by changing an infinite number of element of the sum no? If we change just a finite number, it should be the same, no?

You are right. Changing finite numbers wont change the sum. 
Here is a nice example: Set $S: =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$. Then 
\begin{align*}
\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}-\frac{1}{4n-2}-\frac{1}{4n}\right)
\end{align*}
is a rearrangement of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$ and since
\begin{align*}
\frac{1}{2n-1}-\frac{1}{4n-2}-\frac{1}{4n} = \frac{1}{4n-2}-\frac{1}{4n}=\frac{1}{2}\left(\frac{1}{2n-1} -\frac{1}{2n}\right)
\end{align*}
for all $n\in\mathbf N$ we have
$$\sum_{n=1}^{\infty}\frac{1}{2n-1}-\frac{1}{4n-2}-\frac{1}{4n} =\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2n-1} -\frac{1}{2n}=\frac{1}{2}S.$$
This is just a special case of the Riemann series theorem since $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$ is obviously conditional convergent.
