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In the paper “On the order of magnitude of the difference between consecutive prime numbers” by Harald Cramér there is the following statement:

Suppose $\{X_n\}_{n=2}^\infty$ is a sequence of independent random variables, such that $X_n \sim Bern(\frac{1}{\ln(n)})$. Then $\lim_{n \to \infty} \sup |\frac{\sqrt{\ln(n)}(\Sigma_{i=2}^n X_i - li(n))}{\sqrt{2n \ln(\ln(n))}}| = 1$

However, he does not prove this result there, but rather states, that it is proved in his paper “Prime numbers and probability” (which I could not find)

My question is:

How can this statement be proved?

Probably, it has something to do with the Law of Iterated Logarithm, but I do not know for sure ...

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This is a special case of a general law of the iterated logarithm for non-iid random variables (r.v.'s), which states the following:

Suppose that $Y_1,Y_2,\dots$ are independent zero-mean r.v.'s, $S_n:=\sum_1^n Y_i$, $B_n:=Var\, S_n\to\infty$, $|Y_n|\le M_n\in(0,\infty)$, and $M_n=o((B_n/\ln\ln B_n)^{1/2})$. Then $$\limsup_n\frac{S_n}{\sqrt{2B_n\ln\ln B_n}}=1 $$ almost surely.

See e.g. V. Petrov, Ch. X, Theorem 1. This theorem is due to Kolmogorov (1929).

(Just in case, here is a reference to Cramér's paper:

Cramér, H. 1935 Prime numbers and probability. Skand. Mat.-Kongr. 8, 107--115.

I found it in Granville's paper.)

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