Reversing the order of summation in $\sum_{i=1}^{n} i$

Question

Here is a summation that is supposed to be solved:

$$S_n = \sum_{i=1}^{n} i$$

The author says it can be solved by substituting with $i = n-j$:

$$\sum_{i=1}^{n} i = \sum_{n-j=1}^{n} (n - j)$$

The next step of the simplification he doesn't explain, nor does he explain any simplification past that:

\begin{align*} & = \sum_{j=0}^{n-1} (n-j)\\ & = \sum_{j=0}^{n-1} n - \sum_{j=0}^{n-1}j\\ & = n \sum_{j=0}^{n-1}1 - \sum_{j=1}^{n}j + n \end{align*}

_{This isn't the solution yet I have left that out because it's irrelevant.}

How was he able to change $\sum_{n-j=1}^{n} (n - j)$ to $\sum_{j=0}^{n-1} (n-j)$? Is that a rule of summation that I am not aware of?

Answer 1

All he's doing is reversing the order of summation. Try writing it out. Originally, $i$ ranges from $1$ to $n$, so our summation is: $$ S_n = 1 + 2 + \cdots + i + \cdots + n $$ After making the substitution $i = n - j \iff j = n - i$, we note that $j$ now ranges backwards from $n-1$ to $n-n$, so our summation is: $$ S_n = [n-(n-1)] + [n-(n-2)] + \cdots + [n-(n-j)] + \cdots + [n - (n-n)] $$ By simplifying the terms in the round brackets and reversing the order of summation, we obtain: $$ S_n = [n - 0] + [n-1] + \cdots + [n-j] + \cdots + [n - (n - 1)] $$ so that now $j$ is ranging from $0$ to $n - 1$.

Answer 2

If $i=n-j$, then $\displaystyle \sum_{i=1}^{i=n} i = \sum_{n-j=1}^{n-j=n} (n-j) = \sum_{j=n-1}^{j=0} (n-j) = \sum_{j=0}^{n-1}(n-j)$.