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Since Apéry we know that $\zeta(3)$, where $\zeta$ denotes the Riemann zeta function, is irrational. It is also well known that infinitely many values of the Riemann zeta function at odd positive integers are irrational. Moreover, various results by Zudilin have shown that certain subsets of zeta values at odd positive integers are irrational; for instance, at least one of $ζ(5), ζ(7), ζ(9)$, or $ζ(11)$ is irrational.

Are there any similar results for $P(n)$, where $P$ is the prime zeta function, i.e.,

$$ {\displaystyle P(n)=\sum _{p\,\in \mathrm {\,primes} }{\frac {1}{p^{n}}}={\frac {1}{2^{n}}}+{\frac {1}{3^{n}}}+{\frac {1}{5^{n}}}+{\frac {1}{7^{n}}}+{\frac {1}{11^{n}}}+\cdots ?} $$

A quick search on Wolfram Alpha reveals the following:

I was not able to find any papers or articles related to the irrationality of values of $P$ at positive integers. Have these been studied in a (more or less) serious manner, analogously to $\zeta$? What are the current results?

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    $\begingroup$ Might be a question for MOF $\endgroup$ – timur Oct 17 '18 at 1:29
  • $\begingroup$ @timur I think so too. Once the bounty expires I will post it there. Thank you! $\endgroup$ – Klangen Oct 17 '18 at 7:13
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    $\begingroup$ Now posted to MO, mathoverflow.net/questions/313122/…? $\endgroup$ – Gerry Myerson Oct 18 '18 at 12:46
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Assume that $$ A(x)=\sum_{p\leq x}\frac{\log(p)}{p}=\log x+C+\epsilon(x)\textrm{, }x\rightarrow\infty\tag 1 $$ and $$ \lim_{x\rightarrow\infty}\epsilon(x)=0 $$ Then we set $R(x)=C+\epsilon(x)$ and we have $$ \sum_{p\leq x}\frac{1}{p}=\sum_{p\leq x}\frac{\log p}{p}\frac{1}{\log p}=\frac{A(x)}{\log x}+\int^{x}_{2}\frac{A(t)}{t\log^2(t)}\tag 2 $$ But $A(x)=\log x+R(x)$, where $R(x)=C+\epsilon(x)=C+o(1)$. Using this in (2) we get $$ \sum_{p\leq x}\frac{1}{p}=1+\frac{C}{\log(x)}+\frac{\epsilon(x)}{\log x}+\int^{x}_{2}\frac{dt}{t \log t}+\int^{x}_{2}\frac{C}{t\log^2 t}dt+\int^{x}_{2}\frac{\epsilon(t)}{t\log^2 t}dt $$ But $$ \int^{x}_{2}\frac{dt}{t\log t}=\log\log x-\log\log 2 $$ and $$ \int^{x}_{2}\frac{C}{t\log^2 t}dt=-\int^{x}_{2}d\left(\frac{C}{\log t}\right)=-\frac{C}{\log x}+\frac{C}{\log 2}. $$

It is easy to see someone with the help of De'l Hospital rule, that if also $\epsilon(t)$ continuous in $[2,\infty)$, then $$ \lim_{x\rightarrow\infty} B(x)\log x=\lim_{x\rightarrow\infty} \epsilon(x)-\lim_{x\rightarrow\infty}\frac{\int^{\infty}_{x}\frac{\epsilon(t)}{t\log^2 t}dt}{\frac{1}{\log(x)}}=0. $$ Hence $$ \lim_{x\rightarrow\infty}B(x)\log(x)=0.\tag 3 $$

Hence we get the next

PROPOSITION 1.

If $$ A(x)=\sum_{p\leq x}\frac{\log(p)}{p}=\log x+C+\epsilon(x)\textrm{, }x\rightarrow\infty $$ then $$ \sum^{\infty}_{p\leq x}\frac{1}{p}=\log\log x+C_1+B(x)\textrm{, }x\rightarrow\infty, $$ where $$ C_1=1-\log\log 2+\frac{C}{\log 2}+\int^{\infty}_{2}\frac{\epsilon(t)}{t\log^2 t}dt. $$ and $$ B(x):=\frac{\epsilon(x)}{\log x}-\int^{\infty}_{x}\frac{\epsilon(t)}{t\log^2 t}dt. $$ with $$ B(x)=o\left(\frac{1}{\log x}\right)\textrm{, }x\rightarrow\infty $$

Continuing for the second proposition we have

$$ \sum_{p\leq x}\frac{1}{p^2}=\sum_{p\leq x}\frac{1}{p}\cdot\frac{1}{p}=x^{-1}\sum_{p\leq x}\frac{1}{p}-\int^{x}_{2}\left(\sum_{p\leq t}\frac{1}{p}\right)\left(\frac{-1}{t^2}\right)dt= $$ $$ x^{-1}\sum_{p\leq x}\frac{1}{p}+\int^{x}_{2}\left(\sum_{p\leq t}\frac{1}{p}\right)\left(\frac{1}{t^2}\right)dt= $$ $$ =x^{-1}\left(\log\log x+C_1+B(x)\right)+\int^{x}_{2}\left(\log\log t+C_1+B(t)\right)t^{-2}dt= $$ $$ =x^{-1}\log\log x+C_1x^{-1}+x^{-1}B(x)+\int^{x}_{2}t^{-2}\log\log tdt- $$ $$ -C_1x^{-1}+2^{-1}C_1+\int^{x}_{2}B(t)t^{-2}dt= $$ $$ =2^{-1}C_1+\int^{\infty}_{2}t^{-2}\log\log t dt+\int^{\infty}_{2}B(t)t^{-2}dt+x^{-1}\log\log x+x^{-1}B(x)- $$ $$ -\int^{\infty}_{x}t^{-2}\log\log t dt-\int^{\infty}_{x}B(t)t^{-2}dt.\tag 4 $$ Hence $$ C_2=2^{-1}C_1+\int^{\infty}_{2}t^{-2}\log\log t dt+\int^{\infty}_{2}B(t)t^{-2}dt.\tag 5 $$

And if we set $$ P(x)=x^{-1}B(x)-\int^{\infty}_{x}B(t)t^{-2}dt,\tag 6 $$ in the same way as $B(x)$ (since $\log(x)B(x)$ in null and continuous) $$ \lim_{x\rightarrow\infty}x\log(x)P(x)=\lim_{x\rightarrow\infty}\left(\log(x) B(x)-\log(x)x\int^{\infty}_{x}B(t)t^{-2}dt\right)= $$ $$ =0-\lim_{x\rightarrow\infty}\frac{\log(x)\int^{\infty}_{x}B(t)t^{-2}dt}{1/x}. $$ But with De'l Hospital we have $$ \lim_{x\rightarrow\infty}\frac{\int^{\infty}_{x}B(t)t^{-2}dt}{1/(x\log x)}=\lim_{x\rightarrow\infty}\frac{-B(x)x^{-2}}{-1/(x^2\log^2 x)-1/(x^2\log x)}= $$ $$ =-\lim_{x\rightarrow\infty}B(x)\log x\frac{\log x}{1+\log x}=0. $$ From the above we see that

PROPOSITION 2. $$ \sum_{p\leq x}\frac{1}{p^2}=C_2+x^{-1}\log\log x+B(x)x^{-1}-\int^{\infty}_{x}t^{-2}\log\log t dt+P(x) \textrm{, }x\rightarrow\infty.\tag 7 $$ where $C_2$ is given from (5) and $P(x)$ from (6). Moreover $$ P(x)=o\left(\frac{1}{x\log x}\right)\textrm{, }x\rightarrow\infty $$

Hence we also get the next

THEOREM. $$ \Pi(2)=\sum_{p-prime}\frac{1}{p^2}=C_2, $$ where $$ C_2=2^{-1}C_1+\int^{\infty}_{2}t^{-2}\log\log t dt+\int^{\infty}_{2}B(t)t^{-2}dt, $$ $$ C_1=1-\log\log 2+\frac{C}{\log 2}+\int^{\infty}_{2}\frac{\epsilon(t)}{t\log^2 t}dt $$ and where $\epsilon(n)$ and $C$ is given from the asymptotic expansion $$ \sum_{p\leq x}\frac{\log p}{p}=\log x+C+\epsilon(x)\textrm{, }x\rightarrow\infty. $$ The functions $B(x)$ and $P(x)$ are given from $$ B(x)=\frac{\epsilon(x)}{\log x}-\int^{\infty}_{x}\frac{\epsilon(t)}{t\log^2 t}dt $$ and $$ P(x)=x^{-1}B(x)-\int^{\infty}_{x}B(t)t^{-2}dt $$ It also holds $B(x)=o\left(\frac{1}{\log x}\right)$ and $P(x)=o\left(\frac{1}{x\log x}\right)$, $x\rightarrow\infty$.

The function $E_i(z)$ is given from $$ Ei(z)=-\int^{\infty}_{-z}\frac{e^{-t}}{t}dt. $$

From the above we can say that the value of $\Pi(2)$ can find if we know $C$ and $\epsilon(x)$ of $$ \sum_{p\leq x}\frac{\log(p)}{p}=\log x+C+\epsilon(x)\textrm{, }x\rightarrow\infty. $$ My best knowledge of above formula until today is $$ \sum_{p\leq x}\frac{\log(p)}{p}=\log x+O(1)\textrm{, }x\rightarrow\infty. $$

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  • $\begingroup$ It is not hard to see why the asymptotic for $\sum_{p \le x}\frac{ \log(p)}{p}$ and other sequences (which, once we assume the PNT or the RH, depend only on the Laurent expansion of $P'(s)$ at $s=1$) won't help to obtain $P(2)$ and $P(k)$. $\endgroup$ – reuns Oct 18 '18 at 19:29
  • $\begingroup$ @ renus: Why? Someone can conjecture $\epsilon(x)$ and get evaluations using the C as new constant. The mathematics always work in this way. Also with the help of a nice posed definition many problems can be solved. $\endgroup$ – Nikos Bagis Oct 18 '18 at 20:22
  • $\begingroup$ $P(2)$ doesn't depend directly on the first few Laurent coefficients of $P'(s)$ at $s=1$. $\endgroup$ – reuns Oct 18 '18 at 20:41
  • $\begingroup$ I awarded the bounty for the effort you put into your answer - even though I agree with reuns' comments. $\endgroup$ – Klangen Oct 19 '18 at 9:42

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