Is the concatenation of all prime numbers, a universe/disjunctive number? 
A universe number is a number which contain any finite length string of digits for a base.

Reference here. These are usually called "disjunctive numbers" in English.
Then the question is, is the concatenation of all prime numbers, a disjunctive number?
By the Rosser Theorem we can find that $$\forall n \ge 6,\text{ }\ln(n)+\ln\left(\ln(n)\right)-1<\frac{p_n}{n}<\ln(n)+\ln\left(\ln(n)\right)$$
where $p_n$ is the n-th number prime.
But I don't know what to do after. Any hint would be appreciated.
 A: Yes, using bounds on prime gaps it is possible to show that for any digit string not starting with $0$ there are an infinite number of primes whose decimal representation begins with that string.  It follows that the concatenation of all primes contains every finite digit string.
Furthermore, the real number whose digits correspond to the concatenation of all primes is known as the Copeland–Erdős constant and it is known to be normal in base $10$, so not only does this sequence contain every finite digit string, every string occurs with the same frequency we would expect if it were random.
A: Yes.
From Istrate and Paun we have that, for a strictly increasing sequence of positive integers (such as, trivially, the primes) $a$ such that $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = 1$, and for some base $b > 1$, we may concatenate the base-$b$ expansions of $a$ to produce a disjunctive sequence.
From the prime number theorem, we have $p \sim n \log n$. Therefore $\lim_{n \rightarrow \infty} \frac{p_{n+1}}{p_n} = 1$, where $p_i$ is the $i$th prime. 
Therefore, the sequence produced by concatenating prime integers contains all words, in any base $b > 1$.
See Calude, Priese, and Staiger for a nice introduction to the subject.
