Help to understand $\sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant j \lt k} {\frac{1}{k-j}}$

Question

I got this summation from the book Concrete Mathematics which I didn't exactly understand:

$$ \begin{align} Sn &= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant j \lt k} {\frac{1}{k-j}} \\ &= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant k-j \lt k} {\frac{1}{j}} \\ &= \sum_{1 \leqslant k \leqslant n} \sum_{0 \lt j \leqslant k-1} {\frac{1}{j}} \\ \end{align} $$

I didn't understant why $1 \leqslant j \lt k$ became $1 \leqslant k-j \lt k$ in the second line and why $1 \leqslant k-j \lt k$ became $0 \lt j \leqslant k-1$ in the third line.

Can you guys help me understanding that?

Answer 1

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

Let $k-j =\alpha$. Limits of $\alpha$ will be $1 \leq \alpha \leq k-1$, which is same as $1 \leq k-\alpha \leq k-1$.

$$\implies S_n = \sum_{k=1}^{n}\sum_{k-\alpha=1}^{k-1}\dfrac{1}{\alpha} =\sum_{k=1}^{n}\sum_{\alpha=1}^{k-1}\dfrac{1}{\alpha}= \sum_{k=1}^{n}\sum_{j=1}^{k-1}\dfrac{1}{j}$$

Answer 2

From here

$$S_n = \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant j \lt k} {\frac{1}{k-j}}=\ldots $$

since $k-j$ goes from $k-1$ down to $1$ we have

$$\ldots=\sum_{1 \leqslant k \leqslant n} \, \sum_{1 \leqslant k-j \lt k} {\frac{1}{k-j}} =\ldots$$

now we change name to the index using $j$ insted of $k-j$

$$\ldots=\sum_{1 \leqslant k \leqslant n} \,\sum_{1 \leqslant j\lt k} {\frac{1}{j}}=\sum_{1 \leqslant k \leqslant n} \,\sum_{0 \lt j\leqslant k-1} {\frac{1}{j}}$$