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Let, $$S_n = \sum_{1\leq j<k+j\leq n}^{} 1/k$$

In the book concrete mathematics, The next line shows that,

$$S_n = \sum_{k=1}^n \sum_{j=1}^{n-k} 1/k$$

I am getting very hard times understanding how $1\leq j<k+j\leq n$ implies $1\leq k\leq n$ and $1\leq j\leq n-k$

How is it done? Can you suggest any resource that can help me manipulating inequalities?

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Well, surely $$1\leq j<k+j\leq n$$ implies $1\leq j$. Now substrack $k$ in the inequality and you get $$1-k\leq j-k<j\leq n-k$$ hence $j \leq n-k$. Since $j<k+j$ you get $0<k$ hence $1 \leq k$. Finally $k \leq n-j \leq n.$

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