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Why does $\sum_{n=0}^\infty\left[\sum_{m=0}^n\left[\frac{1}{m!(n-m)!} \cdot A^{n-m} \cdot B^m\right]\right]=\sum_{l=0}^\infty\left[\frac{1}{l!} \cdot A^l\right] \cdot \sum_{m=0}^\infty\left[\frac{1}{m!} \cdot B^m\right]$ hold?

What properties are used here?

In my textbook there is a hint to perform a change-of variables $n=m+l$ and to interchange the summations.

Then we would get: $\sum_{n=0}^\infty\left[\sum_{m=0}^n\left[\frac{1}{m!(n-m)!} \cdot A^{n-m} \cdot B^m\right]\right]=\sum_{m=0}^{m+l}\left[\sum_{m+l=0}^\infty\left[\frac{1}{m!l!} \cdot A^l \cdot B^m\right]\right]$ but now I don't know how to continue, i.e. how to arrive at $\sum_{l=0}^\infty\left[\frac{1}{l!} \cdot A^l\right] \cdot \sum_{m=0}^\infty\left[\frac{1}{m!} \cdot B^m\right]$.

The summand looks fine but how to adapt the summation bounds and why?

Anna D.
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1 Answers1

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We obtain \begin{align*} \color{blue}{\sum_{n=0}^\infty}&\color{blue}{\sum_{m=0}^n\frac{1}{m!(n-m)!}A^{n-m}B^m}\\ &=\sum_{0\leq m\leq n<\infty}\frac{1}{m!(n-m)!}A^{n-m}B^m\tag{1}\\ &=\sum_{m=0}^\infty\sum_{n=m}^\infty\frac{1}{m!(n-m)!}A^{n-m}B^m\tag{2}\\ &=\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{m!n!}A^{n}B^m\tag{3}\\ &\,\,\color{blue}{=\left(\sum_{n=0}^\infty\frac{1}{n!}A^{n}\right)\left(\sum_{m=0}^\infty\frac{1}{m!}B^m\right)}\tag{4} \end{align*} and the claim follows.

Comment:

  • In (1) we rewrite the index range as preparation for the next step.

  • In (2) we exchange the order of summation provided the series converge absolutely, so that reordering does not change the value of the double series.

  • In (3) we change the index of the inner series to start with $n=0$.

  • In (4) we finally write the double series as product of two series.

Markus Scheuer
  • 108,315
  • If a negative factorial ($(n-m)!$ for $n<m$) is considered to be infinite, by expressing it via the Gamma function – G Cab Mar 05 '18 at 21:20
  • @GCab: Here we do not have negative factorials. – Markus Scheuer Mar 05 '18 at 21:22
  • sorry, I misread the upper index of the 2nd sum in first step – G Cab Mar 05 '18 at 21:26
  • @GCab: No problem and thanks for the credit. :-) – Markus Scheuer Mar 05 '18 at 21:27
  • @MarkusScheuer I think I do not understand the second step fully. In the first step you express the double sum using a single summation operator, so nothing has changed there. But what happens exactly in the second step (what is exchanged?) and/or what properties are used? The inner summation I can somewhat understand but why does the outer summation has 'infinity' as upper bound and not 'n' as before (given)? Thanks in advance. – Anna D. Mar 05 '18 at 23:38
  • @Anna: We consider all pairs $(m,n)$ of non-negative integers with the property $0\leq m\leq n<\infty$. This means, that both, $m$ and $n$ fulfill $0\leq m,n<\infty$ with the additional constraint $m\leq n$. In the first series we start summing with $n$ and inner sum over $ m \leq n$. In the other line we start summing with $m$ and inner sum over $ n \geq m$. Two ways to cover a triangular region. – Markus Scheuer Mar 06 '18 at 06:16
  • Consider a simpler region like $0\leq m\leq n\leq 4$. Draw a lattice and consider the admissible pairs $(m,n)$ once row by row and the other time column by column. We exhaust the triangular region once horizontally and the other time vertically. Two out of many different ways to cover all admissible pairs $(m,n)$. – Markus Scheuer Mar 06 '18 at 06:22
  • Thanks for the clarification, I understand it better now. Thumbs up for your elaborate answer, including the comments. Very helpful. – Anna D. Mar 06 '18 at 12:22
  • @Anna: You're welcome and thanks for your nice comment. Good to see the answer is useful. :-) – Markus Scheuer Mar 06 '18 at 12:30