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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.

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closed as unclear what you're asking by Servaes, carmichael561, kingW3, Matemáticos Chibchas, erfink May 14 '17 at 22:40

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    $\begingroup$ "For $r\neq1$, it is easy to see that $\sum_{k=-\infty}^\infty r^k = 0$". Really? Also, the expression above this doesn't make sense unless the sums converge, which they don't. $\endgroup$ – Servaes May 14 '17 at 12:23
  • $\begingroup$ First of all, it is not necessary for a series to converge to have a value. An example would be 1+2+4+8+... = -1. Though it does not make logical sense, it does hold true. This is all part of analytic continuation. $\endgroup$ – Haran May 14 '17 at 17:52
  • $\begingroup$ And lastly, it is actually pretty obvious that $\sum_{k=-∞}^∞ r^k = 0$. We should add $\sum_{k=0}^∞ r^k = \frac{1}{1-r}$ and $\sum_{k=-∞}^{-1} r^k = \frac{1}{r-1}$ . This shows that $\sum_{k=-∞}^∞ r^k = 0$ $\endgroup$ – Haran May 14 '17 at 18:04
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    $\begingroup$ I do not understand any of the sums, not in the least. I can vaguely make sense of some of them in some way, but I want to know what you mean by these sums. If you cannot make precise what you mean, then there is no point in continuing this conversation, as then you do not even remotely have the beginnings of a question. $\endgroup$ – Servaes May 14 '17 at 19:19
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    $\begingroup$ Haran, you might be interested in this discussion of mine : go.helms-net.de/math/divers/ProblemWithBellmatrix.pdf A couple of years I was thinking of a very similar problem finding that assuming it would easily be possible to use that scheme of changing order in a double series runs in very impressive "black holes"... Of course, some of them can be removed. But it needs careful analyses... And I guess your assumption does not work... $\endgroup$ – Gottfried Helms May 16 '17 at 20:34
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(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.

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