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Let $$S(p_n)=\psi(p_{n+1}) - \psi(p_n)$$

where $p_n$ is the $n$-th prime, and $\psi(x)$ second Chebyshev function. With $u=\log(x)/\log(2)$,

This the same as, with the first Chebyshev function $\theta(x):$

$$S(p_n)=\sum_{i=1}^u\theta(p_{n+1}^{1/i})-\sum_{i=1}^u\theta(p_{n}^{1/i})$$

What can be said about $S(p_n)$ bounds, limits, and values?

For example, is this correct:

$$S(p_n)=\sum_{i=p_{n}+1}^{p_{n+1}} \log({i}^{1/i})$$

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with the prime number theorem on mind $ \Psi (x) \to 0 $ so for big primes the different will be about $ \p_{n+1}-p_{n} $ but again in prime number theorem $ p_{n} \to nlog(n) $ so we have that $ S(p_{n}) \to 1+log(n) $ –  Jose Garcia May 30 at 18:15
    
@JoseGarcia I edited the question, please free feel to say more. –  user160140 May 31 at 22:54

1 Answer 1

with the prime number theorem on mind $ \Psi (x) \to 0 $

so for big primes the different will be about $ p_{n+1}-p_{n} $

but again in prime number theorem $ p_{n} \to nlog(n) $ so we have that $ S(p_{n}) \to 1+log(n) $

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