# 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?

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.

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

$$\sum\limits_{j=i}^{n} j = \sum\limits_{j=0}^{n-i} (i+j) = \sum\limits_{j=0}^{n-i} i + \sum\limits_{j=0}^{n-i} j$$

$$n = n + 0 = n + (i-i) = i + (n-i), \ \text{so}$$ $$\sum_{j=i}^n j = i + (i+1) + \ldots + n = (i+0) + (i+1) + \ldots + (i + (n - i)) = \sum_{j=0}^{n-i} (i+j)$$.