Shifting indices of summation How is it possible for one to write $$\sum\limits_{j=i}^{n} j = \sum\limits_{j=0}^{n-i} (i+j) = \sum\limits_{j=0}^{n-i} i + \sum\limits_{j=0}^{n-i} j$$ I still don't understand how the indices were shifted, and why I am not able to do this. Is there a formula? How does one reason about the newly written summation here?
 A: Just write out what the summations mean.
$$\sum\limits_{j=i}^{n} j = i+(i+1)+(i+2)+…+(i+n)$$
$$\sum\limits_{j=0}^{n-i} (i+j) = (i+0)+(i+1)+(i+2)+…+(i+n)$$
$$ = i+(i+1)+(i+2)+…+(i+n)$$
If you think about it for a second, it all becomes clear. Usually, writing examples and noticing a pattern is a good way to go. (This isn’t anything “formal” but it often helps you to get the basic idea.) For example, take the following sum:
$$\sum_\limits{j = 5}^{12} j = 5+6+7+8+9+10+11$$
Imagine you wanted to rewrite the sum by shifting the starting point to, say, $3$.
$$3+4+5+6+7+8+9$$
But all the numbers are shifted down by $2$, so you have to write
$$(3+2)+(4+2)+(5+2)+(6+2)+(7+2)+(8+2)+(9+2)$$
$$= \sum\limits_{j = 3}^{9} j+2$$
You can rewrite the summation in yet another way:
$$ = \underbrace{3+4+5+6+7+8+9}_{j = 3 \text{ to } j = 9}+\underbrace{2+2+2+2+2+2+2}_{{j = 3 \text{ to } j = 9}} = \sum\limits_{j = 3}^{9} j+\sum\limits_{j = 3}^{9} 2$$
which is exactly what the summation means.
$$\sum\limits_{j=i}^{n} j = \sum\limits_{j=i-m}^{n-m} (m+j) = \sum\limits_{j=i-m}^{n-m} m + \sum\limits_{j=i-m}^{n-m} j$$
For shifting down to $0$, you can generalize this as
$$\sum\limits_{j=i}^{n} j = \sum\limits_{j=0}^{n-i} (i+j) = \sum\limits_{j=0}^{n-i} i + \sum\limits_{j=0}^{n-i} j$$
A: $$n = n + 0 = n + (i-i) = i + (n-i), \ \text{so}$$
$$ \sum_{j=i}^n j = i + (i+1) + \ldots + n = (i+0) + (i+1) + \ldots + (i + (n - i)) = \sum_{j=0}^{n-i} (i+j) $$.
A: Let $k=j-i$. When $j=i,k=0$ and when   $j=n,k=n-i$. So we can write $$\sum\limits_{j=i}^{n} j =\sum\limits_{k=0}^{n-i} i+k $$.
Since the summation is independent of change of variable, we can replace the dummy variable $k$ by $j$ itself. Last equation follows from the distributive rule of summation. 
