Convergence/divergence of sum of finite products of functions of reciprocals of natural numbers where the product of the functions equals the identity function.
Yes that was quite the mouthful.
Proposition:
If $$\ \prod\limits_{i = 1}^n {f_i(x)} = x \quad \forall x \in \mathbb{R}^+,\ $$ then
$$\sum\limits_{k = 1}^\infty \left( {\prod\limits_{i = 1}^n {\frac{1}{f_i(a_{ki})}} } \right) $$
converges for some rearrangement $\ (a_{ki})_{i \in {1,...,n }\ and \ k \in \mathbb{N}} \ $ of $\mathbb{N}$, and diverges for some other rearrangement $\ (a_{ki})_{i \in {1,...,n }\ and \ k \in \mathbb{N}} \ $ of $\mathbb{N}$.
Note that I require $a_{ki} \neq a_{k'i'}$ if $k \neq k'$ or $i \neq i'$, i.e. $(a_{ki})$ is a sequence of natural numbers bijective to $\mathbb{N}$.
Is this true?
An example of it working is:
$f_1(x) = f_2(x) = \frac{1}{\sqrt x}$.
An example of divergence: $\frac{1}{\sqrt 1} . \frac{1}{\sqrt 2} + \frac{1}{\sqrt 3} . \frac{1}{\sqrt 4} + \frac{1}{\sqrt 5} . \frac{1}{\sqrt 6} + ... > \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... = \frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... \right)$, which diverges.
An example of convergence: $\frac{1}{\sqrt 1} . \frac{1}{\sqrt 5000} + \frac{1}{\sqrt 2} . \frac{1}{\sqrt 8} + \frac{1}{\sqrt 3} . \frac{1}{\sqrt 27} + \frac{1}{\sqrt 4} . \frac{1}{\sqrt 64} + ... = \frac{1}{\sqrt 5000} + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...,\ $ which converges. I have ignored the repetitions here (for example $\frac{1}{\sqrt 8} . \frac{1}{\sqrt 512}$ isn't a term because that would be a repeat of $\frac{1}{\sqrt 8}$), but I think it's not difficult to modify this "fake sequence" and make it into one that works.
I'm quite sure it works for $n$th roots using AM-GM to show divergence and using similar methods for rearrangements for the square root example I showed above.