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Is there an intuitive explanation for why all the known non-trivial solutions of $$\frac{1}{1^z} - \frac{1}{2^z} + \frac{1}{3^z} - \frac{1}{4^z} + \cdots=0$$ have a real part of $\frac{1}{2}$?

anomaly
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user85798
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    Your premise is incorrect or you forgot a hypothesis. If you look on the Wikipedia page about this function, http://en.wikipedia.org/wiki/Dirichlet_eta_function, you'll see that it's $(1 - 2^{1-s})\zeta(s)$, so it has plenty of explicit zeros with real part not equal to 1/2: any trivial zero of $\zeta(s)$ and also all $s = 1 + 2\pi{i}k/\log 2$ for nonzero integers $k$. – KCd Nov 01 '13 at 11:43
  • I've added 'non-trivial' to it. I'm mainly interested in $0 < Re(s) < 1$ – user85798 Nov 01 '13 at 11:50
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    But in that case your question is equivalent to asking why the Riemann hypothesis is true, since the elementary formula relating that alternating series to the Riemann zeta-function shows the only zeros it has other than zeros of $\zeta(s)$ are zeros of $1 - 2^{1-s}$ besides 1, and the zeros of $1 - 2^{1-s}$ all lie on the line ${\rm Re}(s) = 1$. Where have you seen a list of any zeros of this alternating series that doesn't mention the Riemann hypothesis? – KCd Nov 01 '13 at 11:52
  • ...have you heard of the Riemann hypothesis? – anon Nov 01 '13 at 11:55
  • I'm not really after a proof though, I just want some intuition to what is so special about 1/2 in relation to that alternating series. – user85798 Nov 01 '13 at 11:56
  • There is nothing more special about $1/2$ in relation to the alternating series than in relation to the original Riemann zeta function, so you might as well ask about the Riemann zeta function in particular or $L$-functions in general (making the series alternating is a red herring). And to answer that: a conjectural physical reason to expect $\rm Re=1/2$ is Hilbert-Pólya. (There is a lot of current theoretical work investigating the spectral properties of $L$-functions.) – anon Nov 01 '13 at 11:57
  • I know, but what difference does it make if I'm only concerned with this particular series? - Ok, thanks, I'll read about that – user85798 Nov 01 '13 at 12:09
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    Chapter 3 in this document lists some pieces of "evidence" in favor of the Riemann Hypothesis being true. – Daniel R Nov 01 '13 at 12:11

1 Answers1

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As pointed out in the comments above, your expression is equal to:

$$\frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots = \zeta(s)(1 - 2^{1-s})$$

To illustrate this further we can look at the matrix multiplication that becomes your series above by creating the following two tables:

First table (red and green) called $\zeta(s)$:

$k \mid n : 1$ else $0$.

Second table (red, green and yellow) called $(1 - 2^{1-s})$:

$n=k: 1$ else if $n=2 \cdot k: -2$ else $0$

Zeta(s) and (1-2^(s-1)

Taking the matrix product of the two tables above we get:

Zeta(s) times (1-2^(s-1)

The first column then has the numerators of your series.

Some Mathematica code to try:

Clear[n, s];
s = 2;
Sum[1/(2*n + 1)^s - 1/(2*n + 2)^s, {n, 0, Infinity}]
Zeta[s]*(1 - 2^(1 - s))

which when evaluated gives identical answers.

So because the alternating series has $\zeta(s)$ as a factor, the alternating series must also have all the non-trivial zeros of the $\zeta(s)$ function as zeros.

Edit 22.4.2015: But note that $(1 - 2^{1-s})$ has zeros with real part not equal to $\frac{1}{2}$

Mats Granvik
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