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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?

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2 Answers 2

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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 $n - 1$.

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    $\begingroup$ Why does he want to reverse it? $\endgroup$
    – David G
    Commented Sep 8, 2014 at 22:39
  • $\begingroup$ Presumably, because he's going to do a cool Gaussian pairing trick. Notice that $S_n + S_n$ is: $$ (1 + n) + (2 + (n - 1)) + \cdots + ((n - 1) + 2) + (n + 1) = \underbrace{(n + 1) + (n + 1) + \cdots + (n + 1) + (n + 1)}_{n \text{ times}} = n(n + 1) $$ so it turns out that: $$ S_n = \frac{n(n + 1)}{2} $$ $\endgroup$
    – Adriano
    Commented Sep 8, 2014 at 22:43
  • $\begingroup$ How can I learn to understand this stuff better, like you? I've been reading this stuff for about a week and I'm constantly getting stuck on simple things that I can't think my way out of. $\endgroup$
    – David G
    Commented Sep 8, 2014 at 22:49
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    $\begingroup$ Well that's a bit harder to answer hahah. I guess it just takes a lot of practice and experience. What you're doing is great; you're questioning the steps that the author is making in a proof instead of blindly accepting what they did. Keep up the hard work, and I'm sure that things will start to make more sense. $\endgroup$
    – Adriano
    Commented Sep 8, 2014 at 22:52
  • $\begingroup$ Thanks! Can you also help me understand something else? This is very small and simple: On this page where the author writes $\sum_{k=1}^{n+1} 2k-1=\sum_{k=1}^{n+1} 2k-1+2(n+1)-1$, where is the $\text{...} +2(n+1)-1$ coming from. That's where I got confused. $\endgroup$
    – David G
    Commented Sep 9, 2014 at 18:46
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If $i=n-j$, then $\displaystyle \sum_{i=1}^{i=n} i = \sum_{n-j=1}^{n-j=n} (n-j) = \sum_{j=n-1}^{j=0} (n-j) = \sum_{j=0}^{n-1}(n-j)$.

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  • $\begingroup$ Why does $i=n$ on top of the first summation? $\endgroup$
    – David G
    Commented Sep 8, 2014 at 22:09
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    $\begingroup$ The summation is from $i=1$ to $i=n$. If you make a substitution only in the first value ($i=1$) but not the last value ($i=n$), then you will not be summing the correct terms. So $\sum_{j=0}^{n-1}(n-j)$ in your third display is not correct. $\endgroup$
    – E W H Lee
    Commented Sep 8, 2014 at 22:18

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