2
$\begingroup$

I've got a double sum of the form

$$\sum_{k=0}^\infty \left( \sum_{i=0}^k a_{i,k} \right) $$

and I'm trying to work out how you interchange these two sums. I remember seeing a formula for this in one of my courses, but I can't remember it (nor can I find my notes). As far as I remember, it comes out as two infinite sums.

I know when the inner sum isn't in terms of $k$, you can apply Fubini-Tonelli if the inner summands are all non-negative, but here that obviously doesn't make much sense. I tried to draw a grid with the entries and count them in a different order, but I keep getting the sum indexed by $i$ on the inside so I'm a little lost.

So, if anyone could prod me in the right direction, that would be great.

$\endgroup$
3
$\begingroup$

A good way to remember is to write: $$ \sum_{k=0}^\infty \sum_{i=0}^\infty a_{i,k} \chi(i \le k), $$ where $\chi$ represents the indicator function. Then you simply interchange the summation: $$ \sum_{k=0}^\infty \left( \sum_{i=0}^k a_{i,k} \right) = \sum_{i=0}^\infty \sum_{k=0}^\infty a_{i,k} \chi(i \le k) = \sum_{i=0}^\infty \sum_{k=i}^\infty a_{i,k}.$$ Obviously, you should be concerned with when you can apply Fubini.

$\endgroup$
1
  • $\begingroup$ Thank you! That's a great way of remembering how to do this $\endgroup$ – ribbcastle Dec 13 '16 at 11:57
2
$\begingroup$

A slightly different notation:

\begin{align*} \sum_{k=0}^\infty\left(\sum_{i=0}^k a_{i,k}\right)=\sum_{\color{blue}{0\leq i\leq k<\infty}}a_{i,k}=\sum_{i=0}^\infty\left(\sum_{k=i}^\infty a_{i,k}\right) \end{align*}

$\endgroup$
0
$\begingroup$

Instead of directly switching the sums, you can use an intermediate step with only one sum, for example:

$$\sum_{i=1}^{\infty}\sum_{j=1}^{i}f\left(\left(i,j\right)\right)\\ = \sum_{t\in\left\{ \left(i,j\right)\mid j\le i\le \infty\right\} }f\left(t\right)\\ = \sum_{j=1}^{\infty}\sum_{i=j}^{\infty}f\left(\left(i,j\right)\right)$$

This has the advantage that it works with pretty much arbitrary amounts of nested summations and it's also easier.

$\endgroup$
0
$\begingroup$

You can simply write the terms as

\begin{align} \sum_{k=0}^\infty \sum_{i=0}^k a_{i,k} &= \color{blue}{a_{0,0}}\\ &\ + \color{blue}{a_{0,1}} + \color{green}{a_{1,1}}\\ &\ + \color{blue}{a_{0,2}} + \color{green}{a_{1,2}} + \color{red}{a_{2,2}}\\ &\ + \dots\\ \end{align}

and this suggests reordering in a form \begin{align} \sum_{k=0}^\infty \sum_{i=0}^k a_{i,k} &= (\color{blue}{a_{0,0}}+ \color{blue}{a_{0,1}} + \color{blue}{a_{0,2}}+\dots) \\ &\ + (\color{green}{a_{1,1}} + \color{green}{a_{1,2}} + \color{green}{a_{1,3}} + \dots)\\ &\ + (\color{red}{a_{2,2}} + \color{red}{a_{2,3}} + \color{red}{a_{2,4}} + \dots)\\ &\ + \dots\\ &= \sum_{i=0}^\infty \sum_{k=i}^\infty a_{i,k} \end{align}

$\endgroup$

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.