# Concrete Mathematics Reducing sums to closed form

In reducing the sum $S_n = \sum_{1\leq j<k\leq n}(\frac{1}{k-j})$ to a closed form, the authors start by replacing $k$ with $k+j$, such that $S_n = \sum_{1\leq j<k+j\leq n}(\frac{1}{k})$. The next step then reduces $S_n$ to $\sum_{1\leq k\leq n}\sum_{1\leq j\leq n-k}(\frac{1}{k})$, and giving "summing first on j" as an explanation.

I understand how $[1\leq j < k \leq n]=[1\leq k\leq n][1\leq j < k]=[1\leq j \leq k][j<k\leq n]$ (where $[P]$ denotes the Iverson bracket), which is used in reducing many other sums in the book. But how do the authors use this identity to reduce $\sum_{1\leq j<k+j\leq n}(\frac{1}{k})$ to $\sum_{1\leq k\leq n}\sum_{1\leq j\leq n-k}(\frac{1}{k})$? Is there any other identity used, or the previous one is somehow maniuplated?

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I think the last part of your Iversonian equation should be $[1 \leq j \leq n][j < k \leq n]$ – Dongie Agnir Jul 10 '15 at 1:21

$$k+j\leqslant n\iff\begin{cases}k\leqslant n-j\leqslant n,\quad\text{since }1\leqslant j\\\\j\leqslant n-k\end{cases}$$