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What is the value of $\displaystyle \sum_{n=1}^{\infty} \dfrac{p_i}{i!} =\dfrac{2}{1!}+\dfrac{3}{2!}+\dfrac{5}{3!}+\dfrac{7}{4!}+\cdots$?

This question really interested me since we all know that $\displaystyle \sum_{n=0}^{\infty} \dfrac{1}{n!} = e$. So I would think it would make sense that if we have primes in the numerator it would also converge?

Jacob Willis
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  • At first glance the ratio test will prove convergence. Have you tried that? – hardmath Jan 08 '16 at 04:52
  • Yes, we take $L = \lim_{n \to \infty}\left |\dfrac{a_{n+1}}{a_n} \right |$ and show it is less than $1$. – Jacob Willis Jan 08 '16 at 04:54
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    Are you only asking if it converges, or are you also asking for the value of the sum? Proving convergence is easy. By Bertrand's postulate, there is always a prime between $2^n$ and $2^{n+1}$. Hence, $p_n \le 2^n$, and thus, the sum converges by comparison to $\displaystyle\sum_{n = 1}^{\infty}\dfrac{2^n}{n!} = e^2-1$. Finding a closed form for the summation might not be possible though. – JimmyK4542 Jan 08 '16 at 04:58
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    It converges to something around $4.738638702686$. – Ben Longo Jan 08 '16 at 04:58
  • @guest, that is true but I doubt there is any proof that it is transcendental. Can we at least prove that it is irrational maybe? The standard proof that $e$ is irrational doesn't quite fit since after multiplying by $n!$ we have a first remainder term $\frac{p_{n+1}}{n+1} \gt 1$. – Dan Brumleve Jan 08 '16 at 05:02
  • @DanBrumleve I doubt it: as you observe, following the proof for irrationality of $e$ does not work. Apart from that, simply using the growth rate of the $n$-th prime clearly is not enough; so somehow a more specific property of $p_n$ needs to play a role, and I don't see anything helpful there. – guest Jan 08 '16 at 05:13
  • @JimmyK4542 nice; I was way too hasty to say the OP solved the problem of convergence! – guest Jan 08 '16 at 05:17
  • I am asking for the value of the sum now since we know it must be convergent. – Jacob Willis Jan 08 '16 at 05:18
  • @Jacob, Ben Longo gave an approximation, what else are you looking for? You may be able to find digits of it faster than evaluating the series with some kind of spigot algorithm. – Dan Brumleve Jan 08 '16 at 05:41
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    This is OEIS A$100127$. – Lucian Jan 08 '16 at 11:11
  • @DanBrumleve Can you show me how you know it converges using $L = \left |\dfrac{a_{n+1}}{a_n} \right | <1$? – Jacob Willis Jan 08 '16 at 15:57
  • @guest, I believe its irrationality follows from conjectured bounds on prime gaps (Cramer's conjecture is more than enough but maybe something only slightly stronger than RH works). If infinitely often $\frac{p_n}{n}$ has a very small fractional part, then multiplying the series by $(n-1)!$ will not give an integer, since the following terms will add up to less than $1$. So it's almost the same as the proof for $e$ but it doesn't need to work for every $n$, just an infinite number of them (now I'm not sure if we really need the conjectures...) – Dan Brumleve Jan 08 '16 at 16:01
  • Actually, we don't need the conjectures at all, we don't have to ensure the fractional part is very small, only that it is not too large (less than $1-\frac{\log{n}}{\sqrt{n}}$ or thereabouts), and it should be easy to drum up an infinite supply of $p_n$ that satisfy this. – Dan Brumleve Jan 08 '16 at 16:19
  • @Jacob, rewrite the ratio as $\frac{p_{n+1}}{p_n \cdot (n+1)}$ and use Bertrand's postulate to replace the $\frac{p_{n+1}}{p_n}$ part with an upper bound. – Dan Brumleve Jan 08 '16 at 16:25
  • @DanBrumleve nice, I would not mind seeing this worked out somewhere. Perhaps you can ask and answer a question here to that effect. – guest Jan 08 '16 at 19:52
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    @guest, I posted a question and an answer: http://math.stackexchange.com/a/1605143/1284 – Dan Brumleve Jan 09 '16 at 05:29

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