Let us vertically add $\zeta(k)$ for all integer values of $k$. We would get :
$$\sum_{k=-\infty}^\infty \zeta(k) = \sum_{k=-\infty}^\infty 1^k + \sum_{k=-\infty}^\infty 2^k + \sum_{k=-\infty}^\infty 3^k+\cdots$$
For ($r\neq1$), it is easy to see that $\sum_{k=-\infty}^\infty r^k = 0$.
We know that in these kind of situations, we would derive the same answer at $r=1$ by continuation of the pattern, just like finding $1+2+4+8+\cdots$ is equal to $\frac{1}{1-x}$ at $x=2$ though the value logically only holds at $-1 < x < 1$.
We can even prove that the value is zero at $r=1$.
$$\sum_{k=1}^\infty 1^k = \sum_{k=1}^\infty \frac{1}{k^0} = \zeta(0) = \frac{-1}{2}$$
$$\sum_{k=-\infty}^{-1} 1^k = \sum_{k=1}^\infty \frac{1}{k^0} = \zeta(0) = \frac{-1}{2}$$
$1^k$ at $(k=0) = 1$
Adding the above three would prove that $$\sum_{k=-∞}^∞ 1^k = 0$$. Substituting all of this in the initial equation, we get:
$$\sum_{k=-\infty}^\infty \zeta(k)= 0$$
Is the above correct? Does this give any information about the zeta function that we don't know?
Note: Though some of these series are divergent, they can still be assigned a value through analytic continuation. Take for example $1+2+4+8+...$, though it is divergent, it can be assigned a value of $-1$ through the 2-adic metric. An easy way to see is:
$S=1+2+4+8+...$
$S=1+2(1+2+4+8+...)$
$S=1+2S$
$S=-1$
Any help will be supported. Thank you.
(1) The Dirichlet series only works for $k = \Re k\ge 1$ (the famous $1 + 2 + 3 + \cdots = -1/12$ is utter nonsense).
(2) $\zeta(1) = \infty$.
(3) Rearranging wildly a double sum can lead to more nonsense.