To clarify, I'm asking about series where $n$th term depends only on $n$th prime number. From the famous result (by Euler, I think) we have:

$$\sum_{n=1}^\infty \ln \left(1-\frac{1}{p_n^s} \right)=-\ln \zeta(s)$$

As a particular case:

$$\sum_{n=1}^\infty \ln \left(1-\frac{1}{p_n^2} \right)=-\ln \frac{\pi^2}{6}$$

What about the general series:

$$\sum_{n=1}^\infty f(p_n)$$

Does it have a nontrivial closed form for some other function $f$?

In general, I'm also interested in infinite products (they can be easily written as series, as I did for the titular one), nested radicals, continued fractions, etc.


Thanks to the great advice of @user1952009, I looked up Dirichlet beta function, which has the prime expansion:

$$\sum_{n=2}^\infty \ln \left(1-\frac{(-1)^{(p_n-1)/2}}{p_n^s} \right)=-\ln \beta(s)$$

Among it's special values we have:

$$\sum_{n=2}^\infty \ln \left(1-\frac{(-1)^{(p_n-1)/2}}{p_n} \right)=-\ln \frac{\pi}{4}$$

$$\sum_{n=2}^\infty \ln \left(1-\frac{(-1)^{(p_n-1)/2}}{p_n^2} \right)=-\ln G$$

(Catalan's constant)

$$\sum_{n=2}^\infty \ln \left(1-\frac{(-1)^{(p_n-1)/2}}{p_n^3} \right)=-\ln \frac{\pi^3}{32}$$


  • $\begingroup$ Now I'm curious about $$1+\cfrac1{2+\cfrac1{3+\cfrac1{5+\cfrac1{7+\dots}}}}$$ $\endgroup$ – Simply Beautiful Art Aug 30 '16 at 22:00
  • $\begingroup$ @SimpleArt, no closed form is known. Neither for the nested radical $\endgroup$ – Yuriy S Aug 30 '16 at 22:00
  • $\begingroup$ Darn it, I tried. $\endgroup$ – Simply Beautiful Art Aug 30 '16 at 22:01
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    $\begingroup$ I don't know if this counts, but Euler proved that the sum of the reciprocals of the primes diverges. $\endgroup$ – Simply Beautiful Art Aug 30 '16 at 22:05
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    $\begingroup$ I recall a long time ago proving a series involving primes in the exponent, i'm going to try dig it up, with some luck i might find it $\endgroup$ – frogeyedpeas Aug 30 '16 at 22:24

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