7

I was reading Apostle's Analytic Number Theory book and I saw this formula being used in many cases. Why is this true?

$$ \sum_{n=1}^{\infty} \sum_{d|n} f(d,n) = \sum_{d=1}^{\infty} \sum_{n=1}^{\infty} f(d,nd) $$

I don't see the intuition behind it.

Also, will this hold for finite sums, i.e,

$$ \sum_{n=1}^{m} \sum_{d|n} f(d,n) =^{?} \sum_{d=1}^{m} \sum_{n=1}^{m} f(d,nd) $$

Alex
  • 317
  • Would writing the second sum instead as $\sum_{d=1}^\infty\sum_{r=1}^\infty f(d,rd)$ clarify things? – Angina Seng Nov 07 '18 at 16:55
  • @LordSharktheUnknown So you put $n=rd$? But why does it change the summation index to what it is in the RHS? – Alex Nov 07 '18 at 17:05
  • The intuition is that summing an absolutely convergent series is an "associative" operation: the sum is the same no matter how you "bracket" it. There is actually a notion of "absolute summability", according to which we can write the first sum as $\sum_{n \in \mathbb{N}, \ d \in \mathbb{N}, \ d \mid n} f(d, n)$, and the second as $\sum_{d \in \mathbb{N}, \ q \in \mathbb{N}} f(d, qd)$, and these two sums are equal because there is a bijection between the pairs $(d, n)$ and $(d, q)$. For gory details, see an answer I happen to have just posted: https://math.stackexchange.com/a/2988625/213690. – Calum Gilhooley Nov 07 '18 at 17:08
  • @CalumGilhooley How is there a bijection if $d$ is different in both cases? Also, will this hold for finite sums? – Alex Nov 07 '18 at 17:12
  • 1
    More explicitly and precisely: let $K = {(d, n) \colon d \in \mathbb{N}, \ n \in \mathbb{N}, \ d \mid n}$, and define the pair of mutually inverse bijections $f \colon K \to \mathbb{N}^2$, $(d, n) \mapsto (d, n/d)$ and $g \colon \mathbb{N}^2 \to K$, $(d, q) \mapsto (d, qd)$. It won't work for finite sums. By the way, I agree with Lord Shark that using the same letter $n$ in two different ways in the same equation is confusing, even though it is not strictly incorrect. – Calum Gilhooley Nov 07 '18 at 17:20
  • 1
    I think if you change the upper limit of the summation to $\lfloor{m/d}\rfloor$ in the finite case, it will be OK, but I'm a bit punch-drunk now, so this may be crap! – Calum Gilhooley Nov 07 '18 at 17:28
  • @CalumGilhooley See section 3.10 – Alex Nov 08 '18 at 13:29
  • I may be losing my mind, but I can't see it in section 3.10 of Tom M. Apostol, Introduction to Analytic Number Theory (1976). Have I got the right book? ... I've also checked section 3.10 of Modular Functions and Dirichlet Series in Number Theory (second edition 1990), but I can't see it there, either. – Calum Gilhooley Nov 08 '18 at 14:45

1 Answers1

7

We obtain \begin{align*} \color{blue}{\sum_{n=1}^\infty \sum_{d|n}f(d,n)}&=\sum_{n=1}^{\infty} \sum_{{d=1}\atop{d|n}}^n f(d,n)\tag{1}\\ &=\sum_{{1\leq d\leq n\leq \infty}\atop{d|n}}f(d,n)\tag{2}\\ &=\sum_{d=1}^\infty\sum_{{n=d}\atop{d|n}}^\infty f(d,n)\tag{3}\\ &=\sum_{d=1}^\infty \sum_{{n=d}\atop{dd^{\prime}=n}}^\infty f(d,dd^{\prime})\tag{4}\\ &=\sum_{d=1}^\infty\sum_{d^{\prime}=1}^\infty f(d,dd^{\prime})\tag{5}\\ &\,\,\color{blue}{=\sum_{d=1}^\infty\sum_{n=1}^\infty f(d,nd)} \end{align*} and the claim follows.

Comment:

  • In (1) we write the index range of $d$ more explicitly.

  • In (2) we write the index range somewhat more conveniently.

  • In (3) we change the order of summation.

  • In (4) we introduce $d^\prime$ using the definition of the divisor $d$.

  • In (5) we sum over $d^\prime$ instead of $n$. We observe $d^\prime=1$ if $n=d$, $d^\prime=2$ if $n=2d$, etc.

Similarly we obtain \begin{align*} \color{blue}{\sum_{n=1}^m \sum_{d|n}f(d,n)}&=\sum_{n=1}^{m} \sum_{{d=1}\atop{d|n}}^n f(d,n)\\ &=\sum_{{1\leq d\leq n\leq m}\atop{d|n}}f(d,n)\\ &=\sum_{d=1}^m\sum_{{n=d}\atop{d|n}}^m f(d,n)\\ &=\sum_{d=1}^m \sum_{{n=d}\atop{dd^{\prime}=n}}^m f(d,dd^{\prime})\\ &=\sum_{d=1}^m\sum_{d^{\prime}=1}^{\left\lfloor m/d\right\rfloor} f(d,dd^{\prime})\\ &\,\,\color{blue}{=\sum_{d=1}^m\sum_{n=1}^{\left\lfloor m/d\right\rfloor} f(d,nd)} \end{align*}

Markus Scheuer
  • 108,315
  • I've been in the habit of appealing to Dieudonne, FoMA (5.3.6) to justify the use of infinite unordered sum expressions such as the right hand side of (2), and inferences such as (1) $\implies$ (2) $\implies$ (3). Do you have an alternative reference for this kind of thing? – Calum Gilhooley Nov 08 '18 at 21:56
  • 1
    @CalumGilhooley: This answer uses just algebraic methods treating series formally. A comprehensive introductory treatment of such techniques can be found in chapter 2 Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik. A comprehensive analytic approach is given in Current Topics in Summability Theory and Applications by H.Dutta and B.E. Rhoades (editors). – Markus Scheuer Nov 08 '18 at 23:04
  • From the Amazon preview, it looks as if Dutta and Rhoades use the term absolute summability in this sense - which I infer must be a much better-established sense of the term than the meaning given to it by Dieudonne. The ambiguity is unfortunate! Do you happen to know of any reference to it in that other sense? (I only know section 5.2 of Alan F. Beardon, Limits: A New Approach to Real Analysis (1997), but that doesn't go far. It also takes a rather different approach, which needs some work to reconcile with Dieudonne's.) – Calum Gilhooley Nov 08 '18 at 23:25
  • 1
    @CalumGilhooley: Regrettably, no. I'm not aware of a text using Dieudonne's definition. I think the definition of absolute summability stated in ch. III, Ex. 23 (ii) of Topological Vector Spaces by H.H. Schaefer comes close to it, or close to the characterization given in FoMA 5.3.4. But it seems to be only used in this exercise and nowhere else in the book. – Markus Scheuer Nov 10 '18 at 16:53
  • 1
    Although Schaefer's book is probably a bit advanced for me, it's been on my want list for a long time, so I've ordered a second-hand copy of the (affordable!) first edition. Thank you for the reference. – Calum Gilhooley Nov 10 '18 at 19:05