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Recall $\sum_{n=1}^{\infty}\frac{1}{2^n}=1$.

Denote $\mathbb{P}$ as the set of all prime numbers.

Can one compute $\sum_{p\in\mathbb{P}} \frac{1}{2^p}$, or just approximate?

To generalize, let $C>0$ be some absolute constant, and fix $\alpha \in (0,1)$.

If $\phi:\mathbb{N}\to[0,\infty)$ is an increasing function (sequence), then how "slow" can $\phi$ grow such that the series $\sum_{n=1}^{\infty}C(\alpha)^{\phi(n)}$ converges? I understand the $C$ plays no role in the convergence of the sum, I just like it there.

Obviously, if $\phi$ grows linearly then we are fine, but if $\phi$ is constant then the series diverges. But what if $\phi$ grows logarithmically or some other slow growing function? Any help would be greatly appreciated. Thank you.

VShaw
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    This converges quickly , so a very good numerical approximation is no problem , however I doubt that there is a "closed-form" expression for the value. – Peter Nov 05 '23 at 06:09
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    It's easy to compute the sum, over the primes, of $2^{-p}$; it's $0.01101010001010001010001000001\dots$ – of course, that's in base two. And you get an approximation by cutting it off after a finite number of bits. – Gerry Myerson Nov 05 '23 at 06:29
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    Please don't use one post to ask two very different questions. – Greg Martin Nov 05 '23 at 06:50
  • Related: https://math.stackexchange.com/q/678985/42969 – Martin R Nov 05 '23 at 08:48
  • There is some information about this number at https://oeis.org/A051006 – Gerry Myerson Nov 07 '23 at 03:06

2 Answers2

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$$S_m=\sum _{n=1}^m 2^{-p_n}$$ It would amazing to compute it using the binary representation of the prime numbers.

Nevertheless, it converges very fast to

$$0.41468250985111166024810962215430770836 57742381379169778682\cdots$$ which is the prime constant.

Martin R
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Claude Leibovici
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If $\phi$ grows asymptotically like $c \log(n)$, then $\alpha^{\phi(n)}=\alpha^{c\log_\alpha(n)\log(\alpha)}=n^{c\log(\alpha)}.$ This is just a p-series, and will thus converge if $c\log(\alpha)<-1$, or if $c>-\frac{1}{\log(\alpha)}$. If it grows faster than logarithmic, then it can be bounded by this and will converge, and if it grows slower than it will not converge.

Tristan Liu
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