I am curious if the zeta distribution converges to normal if it is summed over many times. I am particularly curious if this is true for $\zeta$(4). I know that, if it does converge to normal, it would have to be for $\zeta \ge 4$ because the distribution would not have a finite variance otherwise. If I understand the Central Limit Theorem correctly, this distribution would converge if it is IID and has a finite variance. I have tried using the following definition for the CLT to show convergence, but am running into issues. $\lim_{n_\to \infty} P(\sqrt{n}\frac{\frac{S_n}{n} - \mu}{\sigma} \le x) = \Phi (x)$
Where $ S_n = \sum_1^n X_i$ and $\phi (x)$ is the standard normal distribution
and $X_i$ is the zeta distribution.
Since $X_i$ is $\frac{1}{\zeta(4) k^4}$, then $ S_n$ is $\frac{n}{\zeta(4) k^4}$
So, $\lim_{n_\to \infty} P(\sqrt{n}\frac{\frac{S_n}{n} - \mu}{\sigma} \le x) = \Phi (x)$
Reduces down to $\lim_{n_\to \infty} P(\sqrt{n}\frac{\frac{n}{n\zeta(4) k^4} - \mu}{\sigma} \le x) = \Phi (x)$
= $\lim_{n_\to \infty} P(\sqrt{n}\frac{\frac{1}{\zeta(4) k^4} - \mu}{\sigma} \le x) = \Phi (x)$
This diverges as n goes toward infinity. I am uncertain as to what I have done wrong here. Have I made a mistake or does this imply that the zeta distribution does not converge to Normal?
If anyone has some insight into this problem, please let me know.
The CLT is usually stated as $P(\sqrt{n}\frac{\frac{1}{n} S_n - \mu}{\sigma} \le x) \to \Phi(x)$. (Note that the right-hand side is the CDF $\Phi$, not the PDF $\phi$ of the standard normal distribution.) The quantity $\sqrt{n}\frac{\frac{1}{n} S_n - \mu}{\sigma}$ can be rewritten as $\frac{S_n - n\mu}{\sigma \sqrt{n}}$, but is not equal to $\frac{S_n - n\mu}{\sigma/\sqrt{n}}$ as you have written.