TITLE: Donald Knuth's summation notation confuses me.
QUESTION [16 upvotes]: I do not understand a lot of cases of Knuth's summation notation in Concrete Mathematics. In general, I cannot seem to get a grasp on the commutative law as applied to manipulating sums. The commutative law is stated:
$$\sum_{k \in K}a_{k}=\sum_{p(k) \in K}a_{p(k)}$$
$K$ is a set of integers and $p(k)$ is any permutation of the integers in that set.
My biggest issue arises with cases such as this:
$$\sum_{0 \leq k \leq n}(a+bk)=\sum_{0 \leq n-k \leq n}(a+b(n-k))=\sum_{0 \leq k \leq n}(a+bn-bk)$$
How can you justify that the last two expressions are equivalent? I am also struggling with the following example:
$$\sum_{1 \leq j < k \leq n}\frac{1}{k-j}=\sum_{1 \leq j < k+j \leq n}\frac{1}{k}=\sum_{1 \leq k \leq n}\sum_{1 \leq j \leq n-k}\frac{1}{k}$$
To be clear, I understand the substitution. I do not, however, understand the way he expanded the sum and chose those lower bounds and upper bounds. Could anyone possibly explain this notation in the simplest way possible, but no simpler?

REPLY [6 votes]: After a lot of concentration and failure, here's my answer :)
\begin{align}
S_n&=\sum_{1 \leq j < k \leq n}\frac{1}{k-j}\\
\text{Let } k&=j+u\\
S_n&=\sum_{1 \leq j < j+u \leq n}\frac{1}{u}\\
j+u \leq n &\rightarrow j \leq n-u\\
1 \leq j \text{ and } j \leq n-u &\rightarrow 1 \leq j \leq n-u\\
1 \leq n-u &\rightarrow u \leq n-1\\
j < u+j &\rightarrow 0 < u\\
0 < u &\rightarrow 1 \leq u\\
\therefore 1 \leq &u \leq n-1\\
\text{Thus } S_n&=\sum_{1 \leq j \leq n-u}\sum_{1 \leq u \leq n-1}\frac{1}{u}\\
&=\sum_{1 \leq u \leq n-1}\frac{n-u}{u}\\
\frac{n-u}{u}&=0 \text{ at } u=n\\
\therefore S_n&=\sum_{1 \leq u \leq n}\frac{n-u}{u}\\
&=n\sum_{1 \leq u \leq n}\frac{1}{u}-\sum_{1 \leq u \leq n}1\\
&=nH_n-n\\
&=n\left(H_n-1 \right)
\end{align}
(The other sum seems trivial in comparison.)
P.S. I posted this only as a conclusion to the question and demonstrating that I understand the question now due to the help of Math.SE. I also wanted to share my train of thought with the users here.