# Why is this not a valid argument for the Riemann hypothesis?

I am not a trained mathematician, I just enjoy playing around. The Riemann zeta function with the famous Riemann hypothesis has always fascinated me, and as an amateur it's fun to try to manipulate the series that defines this function in different ways.

What I don't understand is why proving the hypothesis is so difficult, when so much is already known about the function. What is wrong with the following argument:

I start from the relationship between the $$\zeta$$ function and the $$\eta$$ function which should be valid on the critical strip:

$$(1-2^{1-s}) \zeta(s) = \eta(s)$$ with $$\eta(s)=\sum_{n=1}^{\infty} (-1)^{n-1}n^{-s}$$

and $$s=a+bi$$. It is known non-trivial zeros should be symmetric about the critical line $$s=1/2$$ in the critical strip. So let's say there are two zeros $$s=1/2+\varepsilon+bi$$ and $$s=1/2-\varepsilon+bi$$ with for convenience $$0\leq \varepsilon<1/2$$. Since the $$1/(1-2^{1-s})$$ factor is never $$0$$, the zeros of the $$\eta$$ and $$\zeta$$ functions should coincide.

Let

$$S_1 = \eta(1/2+\varepsilon+bi) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}n^{-\varepsilon}}{n^{1/2+bi}}$$

and

$$S_2 = \eta(1/2-\varepsilon+bi) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}n^{\varepsilon}}{n^{1/2+bi}}$$

Can we not draw a one-to-one correspondence between terms, i.e. the series are equal (both $$0$$) when all the terms are equal? In this case, this would imply $$n^{-\varepsilon} = n^{\varepsilon}$$, which is only true for $$\varepsilon=0$$, hence all zeros must lie on the critical line?

• if $s=1/2+\varepsilon+bi$ is a zero, then the functional equation says $1-s=1/2-\varepsilon-bi$ is also a zero. Not $1/2-\varepsilon+bi$.
– mike
Aug 29, 2019 at 23:16

No. Two series can of course have the same value without every term being the same. For example, $$2+3-5 = 1+2-3 = 0$$.