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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?

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    The result you are alluding to rather concerns summation procedures such as: take one odd numbered term then two even numbered terms then one odd numbered term again and so on, that is, for the alternating harmonic series you are interested in, $$\sum_{i=0}^\infty\left(-\frac1{2i+1}+\frac1{4i+2}+\frac1{4i+4}\right)$$ The $n$th sum is $$\sum_{i=0}^n\left(-\frac1{2(2i+1)}+\frac1{4(i+1)}\right)=-\frac12\sum_{k=n+2}^{2n+2}\frac1k$$ which happens to converge to $$-\frac12\ln2$$ Other procedures can even yield divergent series. – 2017-02-12
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    @Did: Thank you for your answer. So you agree that changing a finite number of element in the sum won't change the sum ? – 2017-02-12
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    Of course, since for every $n$ after the last element changed, every $n$th partial sum stays the same. – 2017-02-12

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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.