As a part of a proposition about power series that I am trying to prove I have to show that $$ \sum_{m=0}^{\infty} \left(\sum_{n=m}^{\infty} \frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n\right)(x-b)^m = \sum_{n=0}^{\infty}c_n(x-a)^n$$

Proof: What I have done so far is to show that $$\sum_{m=0}^{n} \frac{n!}{m!(n-m)!}(b-a)^{n-m}(x-b)^m = (x-a)^n $$

So the next idea is to use Fubini's rearrangement theorem for infinite series (I have also shown that the series are absolutely convergent). However, the theorem does not accomodate for dependent indices. Intuitively (using triangular summation), I think that the proper way to interchange series would be \begin{equation*} %\begin{array}{ll} \sum_{m=0}^{\infty} \sum_{n=m}^{\infty} \frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m = \sum_{n=0}^{\infty} \sum_{m=0}^{n}\frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m %\end{array} \end{equation*} and thus $$=\sum_{n=0}^{\infty} c_n(x-a)^n$$ However, I do not know how to perform this step rigorously in case of infinite summation.


Using the indicator function

$$\mathbf{1}_{\{m \leqslant n\}}=\mathbf{1}_{\{n \geqslant m\}} = \begin{cases}1 , & n\geqslant m \\ 0, & n < m\end{cases}$$

we have

$$\sum_{m=0}^\infty\sum_{n=m}^\infty f(m,n) = \sum_{m=0}^\infty\sum_{n=0}^\infty f(m,n)\mathbf{1}_{\{n \geqslant m\}}= \sum_{m=0}^\infty\sum_{n=0}^\infty f(m,n)\mathbf{1}_{\{m \leqslant n\}} $$

With absolute convergence we can apply Fubini's theorem to interchange summations and obtain

$$\sum_{m=0}^\infty\sum_{n=m}^\infty f(m,n) =\underbrace{\sum_{n=0}^\infty\sum_{m=0}^\infty f(m,n)\mathbf{1}_{\{m \leqslant n\}}}_{\text{after switching summation order}} = \sum_{n=0}^\infty\sum_{m=0}^n f(m,n)$$

  • $\begingroup$ When will I finally start to do the magic like this on my own ... $\endgroup$ – Cebiş Mellim Jan 7 at 8:35
  • 1
    $\begingroup$ @CebişMellim: Well, your intuition and observation about triangular summation was a good start and it allowed you to reach the correct conclusion. $\endgroup$ – RRL Jan 7 at 17:41

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