# $\psi(p_{n+1}) - \psi(p_n)$?

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 '14 at 18:15
@JoseGarcia I edited the question, please free feel to say more. – user160140 May 31 '14 at 22:54

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 $S(p_{n}) \to 1+log(n)$