# Thread: Sum of a sum.

1. ## Sum of a sum.

Hi:
I want to prove the following identity: $\displaystyle \sum_{i=0}^{m}\sum_{j=0}^{i}f(i,j)=\sum_{j=0}^{m}\ sum_{i=j}^{m}f(i,j)$. I can see this seems to be true tabulating the values of $\displaystyle i$ and $\displaystyle j$ thus:

i j
-------
0 0
1 0,1
... ....
m 0,1,2,...,m

And now, instead of adding over the rows, I can add over the columns. However, I cant prove the identity. Any help will be welcome. Regards.

Note: the table lends itself to the proof. But I'd like to prove the identity by, for instance, manipulating the limits and indexes of the sums or, not so nice, by induction.

2. Originally Posted by ENRIQUESTEFANINI
Hi:
I want to prove the following identity: $\displaystyle \sum_{i=0}^{m}\sum_{j=0}^{i}f(i,j)=\sum_{j=0}^{m}\ sum_{i=j}^{m}f(i,j)$. I can see this seems to be true tabulating the values of $\displaystyle i$ and $\displaystyle j$ thus:

i j
-------
0 0
1 0,1
... ....
m 0,1,2,...,m

And now, instead of adding over the rows, I can add over the columns. However, I cant prove the identity. Any help will be welcome. Regards.

Note: the table lends itself to the proof. But I'd like to prove the identity by, for instance, manipulating the limits and indexes of the sums or, not so nice, by induction.

Perhaps the simplest way to do it is to open up of the sides, using distributivity and associativity, and obtain the other side:

$\displaystyle \sum_{i=0}^{m}\sum_{j=0}^{i}f(i,j)=$ $\displaystyle f(0,0)+\left[f(1,0)+f(1,1)\right]+\left[f(2,0)+f(2,1)+f(2,2)\right]+$$\displaystyle ...+\left[f(m,0)+f(m,1)+...+f(m,m)\right]=$

$\displaystyle [f(0,0)+f(1,0)+...+f(m,0)]+[f(1,1)+f(2,1)+...+f(m,1)]+...+f(m,m)$ $\displaystyle =\sum_{j=0}^{m}\sum_{i=j}^{m}f(i,j)$ Q.E.D.

Tonio

3. or note that first sum assumes that $\displaystyle 0\le i\le m$ and $\displaystyle 0\le j\le i,$ so putting these together we get $\displaystyle 0\le j\le i\le m$ thus $\displaystyle 0\le j\le m$ and $\displaystyle j\le i\le m.$

4. Thank you very much and regards.
Enrique.