1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.