Is this proof valid even though the harmonic series it is based on is divergent?
Prove: $$\sum_{n=2}^\infty (\zeta(n)-1)= 1$$
Where $\zeta$ as in Riemann's Zeta function is summed over all natural numbers -1.
Start with the Harmonic series $$\sum_{n=1}^\infty {1\over n}= 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9... $$
Remove all powers of 2
$ 1/2 +1/4 +1/8+1/16 $ .... sums to 1
Remove all powers of 3
$ 1/3 +1/9 +1/27 $ .... sums to 1/2
Remove all powers of 5
$ 1/5 + 1/25 +1/125 $ ... sums to 1/4
and so on with every series summing to 1 less than the base number since:
$$\sum_{k=1}^\infty 1/N^K = 1/(N-1)$$
We now have the rearranged series
$1 +1 + 1/2 +1/4 +1/5 +1/6 + 1/9 ....$
There is an extra 1 in the beginning of the series while the terms corresponding to 1 less than any power of 2nd degree or higher($ \zeta-1 $) are missing since they were used with the base series. It follows that $$\sum_{n=2}^\infty (\zeta(n)-1)= 1$$
If the proof is valid, can you explain why the divergence of the series does not invalidate it?
\begin{align*} \sum_{n=2}^\infty (\zeta(n)-1) &= \sum_{n=2}^\infty \sum_{k=2}^\infty \frac{1}{k^n} \\ &=\sum_{n=2}^\infty \left( \frac{1}{2^n} + \frac{1}{3^n} + \cdots \right) \\ &=\left( \color{red}{\frac{1}{2^2}} + \color{blue}{\frac{1}{3^2}} + \cdots \right) + \left( \color{red}{\frac{1}{2^3}} + \color{blue}{\frac{1}{3^3}} + \cdots \right) + \cdots \\ &=\left( \color{red}{\frac{1}{2^2}} + \color{red}{\frac{1}{2^3}} + \cdots \right) + \left( \color{blue}{\frac{1}{3^2}} + \color{blue}{\frac{1}{3^3}} + \cdots \right) + \cdots \\ &=\sum_{k=2}^\infty \sum_{n=2}^\infty \frac{1}{k^n} \\ &=\sum_{k=2}^\infty \frac{1/k^2}{1-1/k}\\ &=\sum_{k=2}^\infty \frac{1}{k(k-1)} \\ &= 1 \end{align*}