# Truncated Euler products, Dirichlet eta function, and convergence issues

Can you prove that the following series does not converge if $$\frac{1}{2}<\sigma<1$$, no matter how close to $$1$$ sigma is, and no matter how large $$t>0$$ is? The series is defined as

$$W(\sigma,t)=\sum_{k=1}^\infty \frac{\cos(t\log p_k)}{p_k^\sigma},$$

where $$p_1, p_2,\dots$$ are the prime numbers, with $$p_1=2$$. You can replace $$\log p_k$$ by $$\log k$$ in the cosine function if it is easier (or replace $$p_k$$ by $$k\log k$$), my guess is that in both cases, the two series either diverge simultaneously or converge simultaneously. Divergence (assuming it diverges) would be due to the fact that $$\log p_k$$ or $$\log k$$ (inside the cosine function) grows too slowly. If somehow it can be proved to converge for some $$\sigma<1$$ and some $$t>0$$, then it would have very deep implications regarding the Riemann Hypothesis (see background section below). Replace $$\log p_k$$ by $$k$$ inside the cosine function, and replace $$p_k$$ by $$k$$ in the denominator, and the series converges, according to WolframAlpha, even with $$\sigma=\frac{1}{2}$$.

Of course, using the Dirichlet test, one can prove that the alternate version of the original series converges (the proof may not be trivial), but I am not interested in that case as it is of no help to solve my problem.

For those who don't have the time to read my long argumentation in the next section, they can jump to the conclusion section.

1. Background

I was investigating the following function:

$$\eta_r(s) = \sum_{q\in S_r} (-1)^{v(q)}\frac{1}{q^s},$$

where $$S_r$$ is the set of all strictly positive integers not divisible by any prime strictly larger than $$p_r$$, with $$p_r$$ being the $$r$$-th prime. The smallest element of $$S_r$$ is $$1$$. Here $$v(q) = 1$$ if $$q$$ is odd, otherwise $$v(q)=-1$$. I restricted my analysis to $$s=\sigma +it$$ a complex number with $$\sigma> \frac{1}{2}$$ and $$t>0$$. Note that for the first $$p_r$$ terms in the above series, $$v(q)$$ is perfectly alternating. Indeed, the $$k$$-th term, if $$k\leq p_r$$, is equal to $$(-1)^{k+1}/k^s$$. But after that there are more and more negative terms, dwarfing the positive terms. Yet the series is absolutely convergent because the denominators are growing fast enough, as long as $$r$$ is finite. However, as $$r\rightarrow \infty$$, an issue emerges that ultimately causes $$W(\sigma, t)$$ to diverge. At least this is my understanding (an opinion, not a proof). The connection between $$\eta_r$$ and $$W$$ will be made apparent later in this post.

It is is easy to prove (see note 1 in the Appendix) that

$$\eta_r(s)=(1-2^{1-s})\cdot \prod_{k=1}^r \Big(1-\frac{1}{p_k^s}\Big)^{-1},$$

where $$p_1, \cdots,p_r$$ are the first $$r$$ primes ($$p_1=2$$). Likewise, you define $$\zeta_r(s)$$ by replacing $$v(q)$$ by $$1$$ in the above series defining $$\eta_r(s)$$. It has the same product as $$\eta_r(s)$$ except that the first independent factor is replaced by $$1$$. Unlike the Dirichlet eta function $$\eta(s)$$ corresponding to $$r=\infty$$, here $$\eta_r(s)$$ has no root at $$\sigma=\Re(s)=\frac{1}{2},$$ and a root at $$s=1$$. Just like the $$\eta$$ function, we have

$$\eta_r\Big(1+\frac{2\pi m}{\log 2}\cdot i\Big)=0, m=\pm 1, 2, 3\dots .$$ It has no other roots other than those mentioned, a fact that is easy to prove if $$r$$ is finite (see note 2), and equivalent to RH (the Riemann Hypothesis) if $$r=\infty$$. Indeed, at $$\sigma=\frac{1}{2}$$, even though there is a small domain centered at the origin (it looks like a circle), with no zero in that domain if $$r$$ is finite, that domain shrinks as $$r$$ increases, and becomes empty at $$r=\infty$$, allowing $$\eta_\infty(s)=\eta(s)$$ to have zeroes at $$\sigma=\frac{1}{2}$$ (and as we all know, $$\eta(s)$$ has infinitely many zeroes on the critical line $$\sigma=\frac{1}{2}$$; RH states that these are the only zeroes of $$\zeta$$ in the critical strip).

The natural question that arises is this: what happens when $$r\rightarrow\infty$$? Here I am focusing on $$|\eta_r(s)|^2$$ where $$|\cdot|$$ stands for the modulus. It has the same roots as $$\eta_r$$. It also has a surprisingly simple expression (see note 3):

$$|\eta_r(s)|^2=|1-2^{1-s}|^2\cdot \prod_{k=1}^r \frac{p_k^{2\sigma}}{p_k^{2\sigma}-2 p_k^{\sigma}\cos(t\log p_k)+1}.$$

All factors are positive. Again, $$s=\sigma +it$$. Of course, $$s_0$$ is a zero of $$\eta_r$$ if and only if the above product, evaluated at $$s=s_0$$ is zero. Clearly there are no zeroes except when $$|1-2^{1-s}|=0$$. Note that as $$\sigma=\frac{1}{2}$$ and $$r\rightarrow\infty$$, the infinite product may or may not converge depending on the value of $$t$$. That is, it may vanish even though the finite product does not. That is the case if the cosine term was replaced by $$0$$, and it might explain why something special happens at $$\sigma=\frac{1}{2}$$. Thus, it is quite possible that $$\eta_\infty(s)$$ has some zeroes when $$\sigma=\frac{1}{2}$$. Of course we all know it has infinitely many zeroes when $$\sigma=\frac{1}{2}$$. To avoid this problem, we focus exclusively on $$\frac{1}{2}<\sigma<1$$ and $$t>0$$ (the value $$t=0$$ is another source of problems). Assuming this absence of zeroes (if $$\frac{1}{2}<\sigma<1$$) extends to the case $$r=\infty$$, we are done with proving RH. Unfortunately, something ugly happens at $$r=\infty$$, and as a result, I haven't proved RH.

Now I can make the connection to my original question, and to the series $$W(\sigma,t)$$ defined at the very top of this post. The convergence status of $$|\eta_r(s)|^2$$, as $$r\rightarrow\infty$$, is identical to that of $$|\eta_r(s)|^{-2}$$. We have, after some simple manipulations:

$$|\eta_r(s)|^{-2}=\frac{1}{|1-2^{1-s}|^2} \cdot \Big\{\prod_{k=1}^r \Big(1+\frac{1}{p_k^{2\sigma}}\Big)\Big\} \cdot \Big\{\prod_{k=1}^r \Big(1-\frac{2\cos(t\log p_k)}{p_k^{\sigma}+p_k^{-\sigma}}\Big)\Big\}.$$

The middle product converges if $$\sigma>\frac{1}{2}$$. Based on a theorem (see here) stating that $$\sum a_k$$ and $$\prod (1\pm a_k)$$ have same convergence status if $$\sum a_k^2<\infty$$, the convergence status of the rightmost product is identical that that of the series $$W(\sigma,t)$$ defined at the beginning of my post. Again, we must have $$\sigma>\frac{1}{2}$$ to be able to leverage the theorem in question.

2. Conclusion

I have studied an infinite Dirichlet $$L$$-series $$\eta_r(s)$$ on the critical strip $$\frac{1}{2}<\Re(s)<1$$, based on the first $$r$$ products in the Euler product of $$\zeta$$. The first $$p_r$$ terms are identical to those of the Dirichlet eta series, known to to converge in that strip (unlike the original Riemann zeta series not converging if $$\Re(s)<1$$). Here $$p_r$$ is the $$r$$-th prime, and $$r$$ can be as large as you want. I proved that $$\eta_r(s)$$ has no zeroes in that strip if $$r$$ is finite. I failed to prove that it remains true if $$r=\infty$$ (proving this is equivalent to proving RH). The last hurdle that needed to be overcome is getting a negative answer to the question I asked in this post, that is, the convergence status of $$W(\sigma,t)$$ in the strip in question. I am afraid that the correct answer is that $$W(\sigma,t)$$ diverges in that strip, resulting in my argumentation not being valid anymore at $$r=\infty$$, and thus proving that I did not prove (nor disproved for that matter) RH. There is a very interesting AMS paper on finite Euler products that readers interested in this subject should read. It is entitled Finite Euler products and the Riemann hypothesis. The respected author came close to proving RH on a large portion of the critical strip, see here.

A possible path to get more insights consists of studying ratios such as $$|\eta_r(s)|^2/|\eta_r(s+1)|^2$$ (the denominator has no root if $$\Re(s)>\frac{1}{2}$$) or more complicated ratios of that form.

Update

Another promising approach is to re-scale $$\eta_r$$. For instance, let $$\eta_r^*(s) = \eta_r(s)/\sqrt{\rho_r(s)}$$, with

$$\rho_r(s)=\rho_r(\sigma+it)= \prod_{k=1}^r \Big(1+\frac{2\cos(t\log p_k)}{p_k^{\sigma}+p_k^{-\sigma}}\Big)>0.$$

The product formula for $$|\eta_r^*(s)|^{-2}$$ now converges as $$r\rightarrow\infty$$ (if $$\sigma>\frac{1}{2}$$), making it impossible for $$\eta_\infty^*(s)$$ to have any zero if $$\frac{1}{2}<\sigma<1$$. Indeed,

$$|\eta_r^*(s)|^{-2}=\frac{1}{|1-2^{1-s}|^2} \cdot \Big\{\prod_{k=1}^r \Big(1+\frac{1}{p_k^{2\sigma}}\Big)\Big\} \cdot \Big\{\prod_{k=1}^r \Big(1-\frac{4\cos^2(t\log p_k)}{p_k^{2\sigma}+p_k^{-2\sigma}+2}\Big)\Big\}.$$

3. Appendix

Here I provide proof sketches for some results stated as "obvious" in the main section, and referred to as note 1, note 2, and note 3 in the main section (section 2).

Note 1. Product formula for $$\eta_r(s)$$.

Let $$\zeta_r(s)$$ be defined by the same series as $$\eta_r(s)$$, except that $$(-1)^{v(q)}$$ is replaced by $$1$$. The function $$\zeta_r(s)$$ has the following Euler product representation:

$$\zeta_r(s)=\prod_{k=1}^r \Big(1-\frac{1}{p_k^\sigma}\Big)^{-1} .$$

Proving this is as easy as proving the product representation for the full function $$\zeta$$. Now, thanks to the fact that $$q\in S_r \Rightarrow 2q\in S_r$$, we have:

$$\eta_r(s)=\zeta_r(s)-2\sum_{q\in S_r}\frac{1}{(2q)^s}=\zeta_r(s) - \frac{2}{2^s}\sum_{q\in S_r} \frac{1}{q^s}=\zeta_r(s)-\frac{2}{2^s}\zeta_r(s).$$

Thus $$\eta_r(s)=(1-2^{1-s})\zeta_r(s)$$.

Note 2. Roots of $$\eta_r(s)$$.

Based on the product formula for $$\eta_r(s)$$, the only roots correspond to $$1-2^{1-s}=0$$. This is possible if and only if $$s=1+\frac{2\pi m}{\log 2}\cdot i$$, with $$m=0$$ or $$m=\pm 1, 2, \dots$$.

Note 3. Product formula for $$|\eta_r(s)|^2$$.

We use the product representation for $$\eta_r(s)$$. The modulus of a product is equal to the product of the moduli. Note that $$s=\sigma +it$$. Also,

$$\Big|\Big(1-\frac{1}{p_k^s}\Big)^{-1}\Big|^{2}=\Big|\frac{p_k^s}{p_k^s-1}\Big|^2=\frac{p_k^{2\sigma}}{|p_k^s-1|^2}.$$

Also, $$|p_k^s-1|^2=|p_k^{2\sigma}-2 p_k^\sigma\cos(t\log p_k)+1|$$. Here $$|\cdot|$$ represents the modulus of a complex number on the left side of the equality, and the absolute value of a real number on the right side. To complete the proof, we need to show that the absolute value symbol can be ignored, because the value inside is always positive. Indeed,

$$p_k^{2\sigma}-2 p_k^\sigma\cos(t\log p_k)+1 \geq (p_k^\sigma - 1)^2\geq 0.$$

• Posted ten hours ago, and already up to version six. Apr 26, 2021 at 13:11
• There will be no more edits. Apr 26, 2021 at 16:02

Take some large $$n$$ and consider the summands in $$W(\sigma,t)$$ for which $$t\log p_k$$ lie between $$2n\pi$$ and $$2n\pi+\pi/3$$. These are the primes $$p_k$$ which lie in the interval $$[e^{2n\pi/t},e^{2n\pi/t+\pi/3t}]$$. Writing $$N=e^{2n\pi/t}$$, prime number theorem implies that this interval contains $$\gg\frac{N}{\log N}$$ primes. For each such prime we have $$\cos(t\log p_k)\geq\cos\pi/3$$, and we have $$\frac{1}{p_k^\sigma}\gg N^{-\sigma}$$. Hence the partial sum over the primes in this interval is $$\gg\frac{N^{1-\sigma}}{\log N}$$ which, as $$\sigma<1$$, tends to infinity. This implies that $$W(\sigma,t)$$ doesn't converge.