Close to the integer squares ?? $ f(x) = \sum_{n=1}^{\infty} g(n)^{-x} ,f(4 n) = \zeta(8 n) $

Question

Consider $g(n)$ an integer function mapping positive integers to positive integers.

Also $g(n)$ is strictly nondecreasing. Thus $g(n+1) > g(n) - 1$.

Define $f(x)$ for $x>1$ as

$$ f(x) = \sum_{n=1}^{\infty} g(n)^{-x} $$

Let $ f(4 n) = \zeta(8 n) $ for all integer $n>1$.

Im am interested in solutions $g(n) \neq n^2$.

Are there infinitely many solutions for $g(n)$ ?

Does $g(n)$ stay close to $n^2$ ?

Does $\sum ((g(n) - n^2)^2)^{-1} $ diverge ?

Are there closed form solutions or recursions for some $g(n)$ ?

Is this related to integral transforms ?

What is this type op math called ?

Ofcourse this is a specific case of a large family of similar problems. Therefore general insights are welcome too.

Answer 1

Let $$H(s) = \sum_m g(m)^{-s}-\sum_m m^{-2s}=\sum_{n\ge 1} a(n)n^{-s}$$

Find $N$ the least $n$ such that $a(n)\ne 0$. Then $$\lim_{s\to \infty} N^s H(s)=a(N)$$ This is contradicted by $H(4k)=0$ for all $k$. Thus there is no such $N$ and $$\forall n, a(n)=0, \qquad\sum_m g(m)^{-s}=\zeta(2s)$$

Answer 2

The question can be solved by examining the asymptotic behavior of f(n) and $\zeta(n)$.

We begin by noting that $\lim_{n\to \infty}f(n)=\lim_{n\to \infty}f(4n)=\lim_{n\to \infty}\zeta(8n)=1$, and thus we see that since $g(n)$ is strictly increasing, the only way for the limit to be one would be if the first term were 1, since any other term would go to zero, and thus we get that $g(1)=1$.

We now examine the limiting behavior again, but with O notation. We see that \begin{align} f(4k)-1=&\sum_{n=1}^{\infty}g(n)^{-4k}\\ =&\sum_{n=2}^{\infty}g(n)^{-4k}\\ =&O\left(\frac{1}{g(2)^{4k}}\right)\\ =&\zeta(8k)-1\\ =&\sum_{n=2}^{\infty}n^{-8k}\\ =&O\left(\frac{1}{4^k}\right)\\ \end{align} and thus to make these two sums have the same assympotic behavior we see that $g(2)=4$. We can now repeat this procedure by induction, as seen below:

\begin{align} f(4k)-\sum_{n=0}^{m}\frac{1}{n^{8k}}=&\sum_{n=m+1}^{\infty}g(n)^{-4k}\\ =&O\left(\frac{1}{g(m+1)^{4k}}\right)\\ =&f(4k)-\sum_{n=0}^{m}\frac{1}{n^{8k}}\\ =&\sum_{n=m+1}^{\infty}n^{-8k}\\ =&O\left(\frac{1}{(m+1)^{8k}} \right)\\ \implies& g(m+1)=\frac{1}{(m+1)^2}\\ \end{align}

As you can probably guess, this method can be used to show that lots of infinite sums are equal. This field of math is called asymptotic analysis, but generally, asymptotic analysis is focused on finding the limiting behavior of function, not showing that two functions are different since they have different limiting behaviors.

Hopefully, this is what you are looking for, since I'm just a high school sophomore and I'm very new to Math Stack Exchange.