I am used to the Sigma notation: \begin{align} \sum_{j=1}^{n} \frac{1}{j} \end{align}

I have the following problem to solve:

Let $H_{k}$: \begin{align} \sum\limits_{1 \le j \le k} \frac{1}{j} & = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{k} \tag{1} \\ \sum\limits_{1 \le j \le k} \frac{1}{j} & = H_{k} \tag{2} \end{align} Then: \begin{align} S & = \sum\limits_{1 \le k \le n} \frac{H_{K}}{k} \tag{3}\\ S & = \sum\limits_{1 \le j \le k \le n} \frac{1}{jk} \tag{4} \end{align} Now \begin{align} 2S & = \sum\limits_{1 \le j \le k \le n} \frac{1}{jk} + \sum\limits_{1 \le k \le j \le n} \frac{1}{jk} \tag{5} \\ & = \sum\limits_{1 \le j , k \le n} \frac{1}{jk} + \sum\limits_{1 \le k \le n} \frac{1}{k^{2}} \tag{6} \\ & = H_{n}^{2} + H_{n}^{(2)} \tag{7} \end{align}

I have three questions:

  1. How do we get $(5.)$? Is it a general rule that $\sum\limits_{1 \leq j \leq k \leq n} \frac{1}{jk} = \sum\limits_{1 \leq k \leq j \leq n} \frac{1}{jk}$ or does it just work out in this case? Note: I am talking about the conditions of the Sigmas.
  2. What does $\sum\limits_{1 \leq j , k \leq n}$ mean? I have never seen this notation before. This is my main question.
  3. How do we get from $(5.)$ to $(6.)$? Where does the $\frac{1}{k^{2}}$ come from? (I'll probably be able to understand this after an explanation of 2).

Thanks for the help!

  1. The actual rule is that you can switch $j$ and $k$ everywhere and you will have the same result because they are just dummy variables. The special thing in that step is that $1/(jk)=1/(kj)$.
  2. That means that you sum over all pairs $(j,k)$ with $1 \leq j \leq n$ and $1 \leq k \leq n$.
  3. The point here is that $S$ is the sum over (say) the upper diagonal part of a 2D array (including the diagonal). But the array itself is symmetric, so it's also the sum over the lower diagonal part of the array (again including the diagonal). So if you add $S$ written the first way to $S$ written the second way, then you get the sum over all elements of the array, but with the diagonal double counted. That's what that step says.
  • $\begingroup$ @JMoravitz Thanks for the edit. $\endgroup$ – Ian May 15 at 19:10

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