We know, series $ \displaystyle{\sum_{p \text{ prime}}}\frac{1}{p^{3/2}}$ is Convergent since it is bounded above by the Convergent series $ \displaystyle{\sum_{n=1}^{\infty}}\frac{1}{n^{3/2}}$.

But can we say anything about where both the series converges or any upper bound whose value is known for the previous series ?

Any help would be appreciated. Thanks in advance.

  • $\begingroup$ Dear downvoter, kindly mention the reason to downvote. $\endgroup$ Jun 12, 2020 at 21:51
  • $\begingroup$ (Did not downvote) the latter series sum is Riemann zeta function at $\frac{3}{2}$ (see (9)) $\endgroup$ Jun 12, 2020 at 21:57
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    $\begingroup$ I'm not the downvoter, but it's unclear what you are asking. When you ask "where the series converge," are you asking what they converge to? As for an upper bound, would you be satisfied with $$\sum_{p\text{ prime}}{1\over p^{3/2}}\lt\sum_{n=1}^\infty{1\over n^{3/2}}\lt1+\int_1^\infty{dx\over x^{3/2}}=3$$ $\endgroup$ Jun 12, 2020 at 22:06
  • $\begingroup$ Yeah look I agree with the notion that the identity of actuators of voting events being anon is dumb and pointless or not having a follow up comment enforced as a necessity but I suppose this is something to take up on meta $\endgroup$ Jun 13, 2020 at 0:30
  • $\begingroup$ Thanks @Barry Cipra, I was asking for such an upper bound. $\endgroup$ Jun 14, 2020 at 12:25

1 Answer 1


You ask for the prime zeta function evaluated at $3/2$. $$ P(3/2) = 0.849\,562\,683\,621\,566\,446\,3{\dots} \text{,} $$ which you can replicate yourself. (You also ask for $\zeta(3/2) = 2.612375348685488343{\dots}$ and I discuss a method for evaluating $\zeta$ to high precision in this other Answer on this site.)

Rapid computation of the value of the prime zeta function (implicitly including upper and lower bounds) is described in section 2 of this paper by R.J. Mathar. The sum is broken into the "big part" (terms up to some cutoff), then Moebius inversion (described in the linked Wikipedia article) is used to rewrite the "small part" (terms after the cutoff) as a rapidly converging series. (This is also described in section 2 of this paper by Cohen.)

  • $\begingroup$ Thanks for referencing your answer so thoroughly Eric. $\endgroup$ Jun 13, 2020 at 0:33

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