It's well known that $\zeta(0)=\sum{\frac{1}{n^0}}=\sum{1}=-\frac{1}{2}$, so I know there is something wrong with extending the method Mathologer described here infinetely many times:
Partial sums of $\sum{1}$:
$1, 2, 3, 4, 5, 6...$
Partial averages:
$\frac{2}{2}, \frac{3}{2}, \frac{4}{2}, \frac{5}{2}...$
Partial averages of that:
$\frac{4}{4}, \frac{5}{4}, \frac{6}{4}, \frac{7}{4}...$
It can be easily seen that the partial average after you applied this procedure $n$ times is $\frac{2^n}{2^n}, \frac{2^n+1}{2^n}, \frac{2^n+2}{2^n}, \frac{2^n+3}{2^n}...$
So at any point, the average of the partial sum is $\frac{2^n+c}{2^n}$ with $c \in \mathbb{N}$.
For any $c$ we get $\lim_{n \to \infty}\frac{2^n+c}{2^n}=1$
My question is, where exactly does the error occur? Can't we apply this procedure infinetely many times (but only arbitrarily many times)? If so, why not? Or is the error somewhere else?
There is no error. It is simply the case that the following divergent series:
$$\sum_{k=1}^\infty1$$
is not Cesàro summable.
Secondly,
$$\zeta(0)=-\frac12\ne\sum_{k=1}^\infty1$$
which is a common misunderstanding. If one insists on approaching the problem this way, then you should use the Dirichlet eta function:
$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)$$
where $\eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s}$. Cesàro summing this for $s\le0$ will come out right. For a clear closed form, one should apply the Euler summation formula instead.