Interchanging summation involving divisors in index

Question

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,

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

Answer 1

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*}