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If $f(n)$ is a strictly increasing elementary function from the reals to the reals, and $p(n)$ is the $n$'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?

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What do you mean by closed form, anyway? Do values of the prime zeta function count? – J. M. Nov 3 '11 at 10:53
Also posted to MO, where I expect it to be closed, soon. – Gerry Myerson Nov 3 '11 at 12:57
Clearly, there exist answerers who don't read comments... :D – J. M. Nov 3 '11 at 14:12
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 3 '11 at 15:41
@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 3 '11 at 21:43

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