Symmetry in summation

Question

Can someone please explain the motivation behind this equality?

$\sum_{i=1}^{n} (\sum_{j=1}^{i} + \sum_{j=i}^{n})\frac{1}{ij} = 2\sum_{i=1}^{n} \sum_{j=1}^{i} \frac{1}{ij}$

They explain that it is symmetry of some kind.

Answer 1

To obtain $(1)$, note that $$\sum_{i=1}^n\sum_{j=1}^i\frac{1}{ij} = \sum_{1 \le j \le i \le n} \frac{1}{ij} \overset{\text{(interchange $i$ and $j$)}}{=} \sum_{1 \le i \le j \le n} \frac{1}{ji} \overset{\text{(commutativity)}}{=} \sum_{1 \le i \le j \le n} \frac{1}{ij} = \sum_{i=1}^n\sum_{j=i}^n\frac{1}{ij}$$

Answer 2

We can rewrite the equation as

$$\sum_{i=1}^n\sum_{j=1}^i\frac1{ij}+\sum_{i=1}^n\sum_{j=i}^n\frac1{ij}=2\sum_{i=1}^n\sum_{j=i}^n\frac1{ij}\,,$$

so the question is why

$$\sum_{i=1}^n\sum_{j=1}^i\frac1{ij}=\sum_{i=1}^n\sum_{j=i}^n\frac1{ij}\,.\tag{1}$$

$\sum_{j=1}^i\frac1{ij}$ is the sum of the fractions in row $i$ of the following table, as shown in the last column, and the lefthand side of $(1)$ is therefore the sum of all of the entries in the table, as shown in the lower righthand corner.

$$\begin{array}{c|ccccc|c} i\backslash j&1&2&3&\ldots&n&\\\hline 1&\frac1{1\cdot1}&&&&&\sum\limits_{j=1}^1\frac1{1\cdot j}\\ 2&\frac1{2\cdot1}&\frac1{2\cdot2}&&&&\sum\limits_{j=1}^2\frac1{2j}\\ 3&\frac1{3\cdot1}&\frac1{3\cdot2}&\frac1{3\cdot3}&&&\sum\limits_{j=1}^3\frac1{3j}\\ \vdots&\vdots&\vdots&\vdots&\ddots&&\vdots\\ n&\frac1{n\cdot1}&\frac1{n\cdot2}&\frac1{n\cdot3}&\ldots&\frac1{n\cdot n}&\sum\limits_{j=1}^n\frac1{nj}\\\hline &\sum\limits_{i=1}^n\frac1{i\cdot1}&\sum\limits_{i=2}^n\frac1{i\cdot2}&\sum\limits_{i=3}^n\frac1{i\cdot3}&\ldots&\sum\limits_{i=n}^n\frac1{in}&\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}=\sum\limits_{j=1}^n\sum\limits_{i=j}^n\frac1{ij} \end{array}$$

At the bottom I’ve shown the sums of the columns. Of course the sum of the column sums must be the same as the sum of the row sums, since in both cases we’re just getting the sum of all of the fractions in the body of the table, so the sum across the bottom must be the same as the sum down the righthand edge. The sum across the bottom is

$$\sum\limits_{j=1}^n\sum\limits_{i=j}^n\frac1{ij}\,,$$

and the sum down the righthand edge is

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}\,,$$

so

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}=\sum\limits_{j=1}^n\sum\limits_{i=j}^n\frac1{ij}\,.\tag{2}$$

The actual names of the index variables aren’t important: clearly

$$\sum\limits_{j=1}^n\sum\limits_{i=j}^n\frac1{ij}=\sum\limits_{\ell=1}^n\sum\limits_{k=\ell}^n\frac1{k\ell}\,,$$

where we’ve renamed $i$ to $k$ and $j$ to $\ell$. Thus, $(2)$ says the same thing as

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}=\sum\limits_{\ell=1}^n\sum\limits_{k=\ell}^n\frac1{k\ell}\,.\tag{3}$$

Similarly, we can rename $k$ to $j$ and $\ell$ to $i$ and see that

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}=\sum\limits_{\ell=1}^n\sum\limits_{k=\ell}^n\frac1{k\ell}=\sum\limits_{i=1}^n\sum\limits_{j=i}^n\frac1{ij}\,,$$

which is exactly what we wanted back at $(1)$.