Got confused by the statement as the author does not provide any details regarding this property. So in the book "Concrete Mathematics" the authors state:
An index variable that doesn’t appear in the summand (here j) can simply be eliminated if we multiply what’s left by the size of that variable’s index set (here $n-k$).
Then the authors use this property in evaluating the particular sum:
$ \begin{equation} S_n = \displaystyle\sum\limits_{1 \le j < k \le n}^{}{\frac{1}{k-j}} \end{equation} $ = $ \begin{equation} S_n = \displaystyle\sum\limits_{1 \le j < k+j \le n}^{}{\frac{1}{k}} \end{equation} $ - replacing $k$ by $k+j$
$ \begin{equation} S_n = \displaystyle\sum\limits_{1 \le k \le n}^{} \displaystyle\sum\limits_{1 \le j \le n-k}^{}{\frac{1}{k}} \end{equation} $ -summing first on j
$ \begin{equation} S_n = \displaystyle\sum\limits_{1 \le k \le n}^{}{\frac{n-k}{k}} \end{equation} $ - the sum on $j$ is trivial
I have underscored what makes me so confused. According to the authors' note, this is valid, however I seek for the clear explanation or proof that allow to eliminate the index variable by simply multiplying the summand by the upper bound. Could anyone shed more light on that?
In the last line the following general rule is used:
$$\sum_{j=1}^m c_k = \underbrace{c_k+\cdots + c_k}_{m \times c_k} = mc_k$$
Nevertheless, there is a mistake in the index $k$. When replacing summation indices it is better not to use the same index as it happened in your book:
So, let $d:= k-j$. Then,
$$1\leq d \leq \color{blue}{n-1} \text{ and } 1 \leq j \leq n-d$$
It follows
$$S_n = \displaystyle\sum\limits_{1 \le j < k \le n}^{}{\frac{1}{k-j}} = \sum_{d=1}^{n-1}\sum_{j=1}^{n-d}\frac 1d = \sum_{d=1}^{n-1}\frac{n-d}{d}$$