Multiple sum variable notation

Question

I have a question regarding multiple sum variables. Is it correct to rewrite this sum like this ?

$$ \sum_{1\leq i<j\leq n}^{}\frac{i}{j} = \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{i}{j} $$

and how do we rewrite sums like this in general ?

Answer 1

It takes experience and I still make mistakes. What helps is trying small $n$ (1, 2, 3 are usually enough) and seeing if the sums are the same.

The outer sum has to involve $n$, since it is done first and the inner sum done for each value in the inner sum.

The sum can be written in (at least) two ways.

$\begin{array}\\ \sum_{1\leq i<j\leq n}^{}\frac{i}{j} &= \sum_{i=1}^{n}\sum_{j=i+1}^{n}\frac{i}{j}\\ &= \sum_{j=1}^{n}\sum_{i=1}^{j-1}\frac{i}{j}\\ \end{array} $

Since the inner sum is empty for $i=n$ in the first sum and $j=1$ for the second sum, they can also be written like this.

$\begin{array}\\ \sum_{1\leq i<j\leq n}^{}\frac{i}{j} &= \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{i}{j}\\ &= \sum_{j=2}^{n}\sum_{i=1}^{j-1}\frac{i}{j}\\ \end{array} $

Since a sum is zero when the lower bound is greater than the upper bound, the two ways are the same.

You can also pull out factors that do not depend on the inner index of summation, so the sums could be written like this.

$\begin{array}\\ \sum_{1\leq i<j\leq n}^{}\frac{i}{j} &= \sum_{i=1}^{n}\sum_{j=i+1}^{n}\frac{i}{j}\\ &= \sum_{i=1}^{n}i\sum_{j=i+1}^{n}\frac{1}{j}\\ \text{and}\\ &= \sum_{j=1}^{n}\sum_{i=1}^{j-1}\frac{i}{j}\\ &= \sum_{j=1}^{n}\frac1{j}\sum_{i=1}^{j-1}i\\ \end{array} $

Answer 2

The second sum defines $j$ in terms of $i$. You can't also have the first sum define $i$ in terms of $j$. That would just take your round in a circle.

We sum over all pairs $(i,j)$ such that $1\leq i<j\leq n$ What values can $i$ take? The smallest is $1$, obviously, and the largest is $n-1$, because $j$ can't be bigger than $n$ and $i$ must be less than $j$. That accounts for the indices on the first sum. You've already figured out how to write the indices on the second sum.

Answer 3

The argument in the sum can (and should) be ignored here. If you regard the coordinate $i$ as indicating the rows and $j$ the columns, of an $n\times n$ matrix $(a_{ij})$ then $\sum_{1\leq i<j\leq n}a_{ij}$ is a finite sum over the matrix elements above the off-diagonal of the matrix (note, $j$ starts at $2$). Since the sum is finite, you can do it any way you want. Now, fix a column $j\ge 2$. Then, the sum of the elements in the $j^{\text{th}}$ column are those from the first to the $(j-1)^{\text{th}}$ row. That is, $\sum^{j-1}_{i=1}a_{ij}.$ Now, sum over all such columns. There are $n-1$ of them, and so you get $\sum^n_{j=2}\sum^{j-1}_{i=1}a_{ij}.$

Answer 4

You should rearrange it as $\sum_{i=1}^{n-1}\sum_{j=i+1}^n\frac{i}{j}$ or $\sum_{j=2}^n\sum_{i=1}^{j-1}\frac{i}{j}$; either way, you have an outer sum with an index whose limits are functions of $n$, and an inner sum with an index whose limits are functions of $n$ and the outer sum's index.