p-i-ith-prime-let-p-n-p-n-1-is-lim-n-

Question Number 3659 by Filup last updated on 18/Dec/15

p_{i} = i th prime

let : ρ = p_{n} - p_{n - 1}

is :

\lim_{n \to \infty} ρ = \infty ?

Commented by Filup last updated on 18/Dec/15

I think it is not . If n \to \infty,

(p_{n - 1} < p_{n})

p_{n} \to \infty \land p_{n - 1} \to \infty

\infty - \infty \neq \infty (it is undefined)

as n \to \infty

p_{n - 1} \to p_{n}

p_{n} \to \infty

∴ \lim_{n \to \infty} ρ < \infty

objections ?

Commented by prakash jain last updated on 18/Dec/15

prime gap g (n)

Prime g (n) ≫ \frac{(\log n) (\log \log n) (\log \log \log \log n)}{\log \log \log n}

so \lim n \to \infty g (n) should be infinity .

Commented by 123456 last updated on 18/Dec/15

what about twins prime

its look like

\underset{n \to \infty}{\lim s u p} ρ = \infty

\underset{n \to \infty}{\lim i n f} ρ = 2 (if exists infinite twins prime)

Commented by prakash jain last updated on 18/Dec/15

It has been proven .

\underset{n \to \infty}{\lim i n f} ρ < 7 \cdot 10^{7}

It only means that are infinitely many ρ

that are less than 7 \cdot 10^{7} .

What is the value of limit on the lower bound

the function that I took . I did not calculate

the limit just assumed it .

Commented by prakash jain last updated on 18/Dec/15

The limit of the lower bound function is

infinity .

Commented by prakash jain last updated on 18/Dec/15

So we have ρ (n) < 7 \times 10^{7} and greater than

\frac{(\log n) (\log \log n) (\log \log \log \log n)}{\log \log \log n} \to \infty

for infinitly many values of n .

So the limit really does not exist .