Why does the expansion of the Riemann zeta function for $\zeta(3)$ to $\sum_{n=1}^{\infty}\frac{n^3}{n^6}$ not allow for substitution of the latter?

Question

For the Riemann-zeta function $\zeta(s)$ can we represent $\zeta(3)$ as the following? $$\zeta(3)=\sum_{n=1}^{\infty}\frac{1}{n^3}=\sum_{n=1}^{\infty}\frac{1}{\biggl(\frac{n^6}{n^3}\biggr)}=\sum_{n=1}^{\infty}\frac{n^3}{n^6}$$

Because $$\zeta(6)=\sum_{n=1}^{\infty}\frac{1}{n^6}=\frac{\pi^6}{945}$$

is it appropriate to write this as

$$\zeta(3)=\sum_{n=1}^{\infty}\frac{n^3}{\biggl(\frac{\pi^6}{945}\biggr)}=\sum_{n=1}^{\infty}\frac{945n^3}{\pi^6}$$

This answer just diverges, so what am I missing? If I take out the constant to yield $$\zeta(3)=\frac{945}{\pi^6}\sum_{n=1}^{\infty}n^3$$ the answer still diverges.

Answer 1

Unfortunately, the sum operation is not multiplicatively distributive. $$\sum ab\neq \sum a \cdot \sum b$$ What you can do is

$$\sum_{n\geqslant 1} \dfrac{1}{n^3}=\sum_{n=m\geqslant 1}\dfrac{m^3}{n^6}=\left[\sum_{n,m\geqslant 1}-\sum_{n\neq m\geqslant 1}\right]\dfrac{m^3}{n^6}=\left[\sum_{n,m\geqslant 1}-\sum_{n> m\geqslant 1}-\sum_{m> n\geq 1}\right]\dfrac{m^3}{n^6}$$ of which, the first part is

$$\zeta(6)\sum_{m\geqslant 1} m^3.$$