Question

Playing around

Answer 1

I found an upper bound for this, but I'm looking for a smaller one, or maybe even an approximation. Just playing around with primes in general if anyone wants to play.

\[\sum_{n=1}^\infty \frac{1}{(p_{n+1}-p_n)^n}\]

Answer 2

\(2,3,5,7,\ldots \)

For \(n\ge 2\) :
\[p_{n+1} - p_n \ge 2 \implies \dfrac{1}{p_{n+1} - p_n} \le \dfrac{1}{2}\]


\[\sum_{n=1}^\infty \frac{1}{(p_{n+1}-p_n)^n} \\~\\ = \dfrac{1}{3-2} + \sum_{n=2}^\infty \frac{1}{(p_{n+1}-p_n)^n}\\~\\ \le \dfrac{1}{3-2} + \sum_{n=2}^\infty (1/2)^n \\~\\\le 1 + \dfrac{1}{2}\]

Answer 3

Pretty sure you have that already..

Answer 4

Yeah, that's basically what I was playing with, I was wondering if it might be possible now to tighten the bound instead of just looking at primes all being odd, or equivalently congruent to 1 mod 2, can we tighten the upper bound on this by extending it to 1 and 5 mod 6?

I was thinking we could assume something "like" this, but I know this is explicitly wrong but maybe you get my idea,

\[\sum_{n=1}^\infty \frac{1}{(p_{n+1}-p_n)^n} < \sum_{n=1}^\infty \frac{1}{2^{2n}} + \sum_{n=1}^\infty \frac{1}{4^{2n+1}}\]

Basically what I'm trying to do is repeat like we did to make the previous upper bound of essentially "assuming all primes are odd numbers" except now I'm "assuming all primes are 1 and 5 alternating" to create the new upper bound. I'm not really sure if this works/or how to do this though.

Answer 5

Nice, I think it is fine to split like that as the series is absolutely convergent.

\[\sum_{n=1}^\infty \frac{1}{(p_{n+1}-p_n)^n} = \sum\limits_{n=1}^{\infty}\dfrac{1}{(p_{2n}-p_{2n-1})^{2n-1}} + \dfrac{1}{(p_{2n+1}-p_{2n})^{2n}}\]

Answer 6

Yeah! This is where I am getting stuck though cause I don't think the primes line up so nicely like this. How about this,

\[p_n \equiv 5 \mod 6\]\[q_n \equiv 1 \mod 6\]
This means:
\[\Delta p_n \ge 2\]\[\Delta q_n \ge 4\] so like:

\[\sum_{n=1}^\infty \frac{1}{(\Delta P_n)^n} = \frac{1}{3-2} + \frac{1}{(5-3)^2} + \sum_{n=1}^\infty \frac{1}{(\Delta p_n)^{\alpha(n)}} + \sum_{n=1}^\infty \frac{1}{(\Delta q_n)^{\beta(n)}}\]

Hopefully the notation is sort of self-explanatory. If not, I can definitely explain better but basically I am not sure how to move forward but I have some ideas. I don't wanna go into them though cause I just dropped a lot of new notation down haha.

Answer 7

Ahh I see. Since 1,5 mod 6 primes occur randomly, relating them with the index n is kinda tricky.

Earlier, I was considering 3 consecutive primes and trying to use the fact that the closest they can get is (p, p+2, p+6)
https://en.wikipedia.org/wiki/Prime_triplet

Answer 8

\[\sum_{n=1}^\infty \frac{1}{(p_{n+1}-p_n)^n} \\~\\= \sum\limits_{n=1}^{\infty}\dfrac{1}{(p_{2n}-p_{2n-1})^{2n-1}} + \dfrac{1}{(p_{2n+1}-p_{2n})^{2n}}\\~\\

\le \dfrac{1}{3-2} + \dfrac{1}{(5-3)^2} +\sum\limits_{n=2}^{\infty} (1/2)^{2n-1} + (1/4)^{2n}\\~\\
= \dfrac{1}{3-2} + \dfrac{1}{(5-3)^2} + \dfrac{1}{6} + \dfrac{1}{240}
\]

Answer 9

I want to agree but something about it seems off to me since I think it's assuming a certain order or something. I like these prime triplets thing, I think that could come in handy. I think maybe I just am not in the right frame of mind for this problem right now maybe I'll come back to it tomorrow.

Answer 10

I think we may consider "n" consecutive primes and further lower the upper bound

Answer 11

grouping more consecutive primes gives us more multiples : 3, 5 etc