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Let $f(n)$ be a strictly increasing elementary function from the reals to the reals, and let $p(n)$ denote the $n^{\rm th}$ prime number. Is there any $f(n)$ such that $\sum_{n=1}^\infty\frac{1}{f(p(n))}$ is algebraic, or has a closed form in terms of elementary functions?

If not, is there a method to prove that the sum must be transcendental?

Klangen
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TROLLHUNTER
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    What do you mean by closed form, anyway? Do values of the prime zeta function count? – J. M. ain't a mathematician Nov 03 '11 at 10:53
  • @J.M. http://en.wikipedia.org/wiki/Closed-form_expression No limits, infinite sums or integrals. Values of prime-zeta are not closed form if I read that article correctly. – TROLLHUNTER Nov 03 '11 at 10:56
  • "...if it can be expressed analytically in terms of a bounded number of certain 'well-known' functions." - and prime zeta is pretty well-known, so... if you meant to ask if it can be expressed in terms of elementary functions, then we're cooking... my point being: "closed form" is relative! See this. – J. M. ain't a mathematician Nov 03 '11 at 11:02
  • Right, bounded number of elementary functions – TROLLHUNTER Nov 03 '11 at 11:03
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    Also posted to MO, http://mathoverflow.net/questions/79904/convergent-series-with-primes where I expect it to be closed, soon. – Gerry Myerson Nov 03 '11 at 12:57
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    Clearly, there exist answerers who don't read comments... :D – J. M. ain't a mathematician Nov 03 '11 at 14:12
  • Which condition are you willing to relax? Is $Im(f) \subset (0, \infty)$ good or not? – N. S. Nov 03 '11 at 15:30
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    Then, if you also include the inverse trig functions in the "elementary" functions list, I think that $f(n) = \frac{1}{ \ln (1- \frac{1}{n^2})}$ will do. Your series then becomes the ln of the Euler product. Of course $\ln( \frac{\pi^2}{6})$ is probably transcendent, but it can be expressed an combo of logs and inverse trig... And the logarithm can be any base.... Oh, wait $f(1)$ doesn't make any sense, the function is defined only from $n=2$. Probably someone can fix it though... :( – N. S. Nov 03 '11 at 15:41
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    @user9176: $f(1)$ doesn't have to make sense; only $f(p(1)) = f(2)$, ne? I would write that up as an answer since it seems to clearly answer the broader (elementary) version of the question; now I'm left wondering about the algebraic version of this... – Steven Stadnicki Nov 03 '11 at 21:43
  • By the way, there will likely not be a way to prove the sums are transcendental (even if it was true). In fact, a proof of transcedence for even a small class of functions would probably require a big leap forward over current methods. – B R Nov 06 '11 at 19:36

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