Does the following 'alternating' Euler product, with $p_n$ the $n$-th prime number, converge for $\Re(s)>0$ ?

$$\displaystyle \prod_{n=1}^\infty \left( \dfrac{1}{1-\frac{1}{p_{n}^{s}}} \right)^{(-1)^n}$$

Based on numerical evidence, I dare to conjecture that this is indeed the case (note that I could not find any zeros), but keen to find approaches towards a proof. Note that I also tried other triggers than $n$ to 'flip the factors', like for instance prime congruence to either $p_n \pmod 6 =1$ or $5$ and $p_n \pmod 4 =1$ or $3$. Even tried the Möbius function $\mu(n)$ as the exponent of $(-1)$, but did not observe any convergence in the domain $\Re(s)\le 1$. So, using $n$ to flip any other prime factor, seems a delicate choice to make convergence work or fail in this domain.


This question loosely builds on: Equality of an alternating infinite product and an infinite sum


3 Answers 3


Taking logarithms, we arrive at the sum

$$-\sum_{n=1}^{\infty} (-1)^n \log(1-p_n^{-s}).$$ For $Re(s)>1/k$, we can approximate the logarithm with its taylor series to $2k+1$ terms with an error of $o(p_n^{-2})$ which converges absolutely. Thus, it suffices to show that for each $s>0$, the sum

$$\sum_{n=1}^{\infty} (-1)^n p_n^{-s}$$ converges. To do this, we estimate

$$p_{2n}^{-s}-p_{2n+1}^{-s} = s^{-1}\int_{p_{2n}}^{p_{2n+1}} x^{-1-s} dx = O_s(G_{2n}\cdot p_{2n}^{-1-s})$$ where we define $G_n= p_{n+1}-p_n$.

Now note that by the PNT we have $p_n\sim n\log n$ and $\displaystyle\sum_{n=X}^{2X-1} G_n = p_{2X}-p_X \sim X\log X$.

\begin{align*} \sum_{n=1}^{\infty} (-1)^n p_n^{-s}&=\sum_{n=1}^{\infty} (p_{2n}^{-s}-p_{2n+1}^{-s})\\ &\ll_s \sum_n G_{2n}\cdot p_{2n}^{-1-s}\\ &< \sum_n G_n \cdot p_n^{-1-s}\\ &\ll \sum_{m=0}^{\infty} \sum_{n=2^m}^{2^{m+1}-1} G_n (m2^m)^{-1-s}\\ &\ll \sum_{m=0}^{\infty} (m2^m)^{-s}\\ \end{align*}

Which converges for $s>0$.

  • $\begingroup$ Many thanks, Jacob. I honestly had not expected a proof being within reach at all. Does this also immediately prove the similar convergence of $\prod_{n=1}^\infty \left( \dfrac{1}{1-\frac{1}{n^{s}}} \right)^{(-1)^n}$, since the first two steps in your proof would be similar and lead to proving that $\sum_{n=1}^{\infty} (-1)^n n^{-s}$ (i.e. the Dirichlet Eta-function) converges for $\Re(s) > 0$ (which already has been proven)? $\endgroup$
    – Agno
    Mar 6, 2014 at 18:24
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    $\begingroup$ Yup. In fact, you can prove $$\prod_{n=1}^{\infty}(1-a_n^{-s})^{(-1)^n}$$ converges for $Re(s)>0$ for any increasing sequence of positive integers $a_n$ this way. The PNT thing I used was overkill. $\endgroup$
    – jacob
    Mar 9, 2014 at 6:57
  • $\begingroup$ Thanks and understood. For the alternating prime product, I had of course hoped for convergence to known constants or to find some new zeros, but did not find any. However, for the alternating product over integers: $\prod_{n=1}^\infty \left( \dfrac{1}{1-\frac{1}{n^{s}}} \right)^{(-1)^n}$, there are a few interesting outcomes. E.g. $s=1$ gives $\frac{\pi}{2}$ and $s=2$ yields $\frac{\pi^2}{8}$. I guess there also exist more complicated closed forms for $s=3,4,5...$. $\endgroup$
    – Agno
    Mar 9, 2014 at 17:40

As Jacob mentioned, the question comes down to the conditional convergence of $\sum p^{-s}(-1)^n$ (for essentially the reason that $\prod(1-x)$ converges and is nonzero if $\sum x$ converges and $\sum x^2$ absolutely converges.

Now, if you look in Titchmarsh's Theory of Functions, there is a theorem that the region of conditional convergence of a Dirichlet series is a half plane (along with part of its boundary). This is proven by showing that convergence at some $s$ gives convergence on every wedge with vertex at $s$ and otherwise strictly to the right.

Thus since the series converges for every real $x>0$ (alternating series), it also converges in the right half plane.


Combine two consecutive terms as follows: $$ \left(1-\frac{1}{p_{2k}^s}\right)\left(1-\frac{1}{p_{2k+1}^s}\right)^{-1} = \frac{p_{2k+1}^s(p_{2k}^s-1)}{p_{2k}^s(p_{2k+1}^s-1)} = 1+\frac{p_{2k+1}^s-p_{2k}^s}{p_{2k}^s(p_{2k+1}^s-1)} $$ Since $\sum\frac{1}{p_{2k}}=\infty$ the sum in product in question tends to $\infty$ as $s\rightarrow\frac{1}{2}$ from the right. Hence the product does not converge for any $s\in[0, \frac{1}{2}]$.

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    $\begingroup$ Note that the numerator $p_{2k+1}^{s}-p_{2k}^s$ can also be small. I don't think what you say holds. $\endgroup$
    – Lucia
    Mar 6, 2014 at 15:11

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