I show that p_(s+k)
< 2*p_(s+k-n) for all positive integers k. Where p_s is the n-th
Ramanujan prime, so R_n = p_s. If you let i = s+k, the
inequality becomes p_i < 2 p_(i-n).

Note that with k=0, we have Lemma
1.i, and note that p_pi(R_n/2) = p_pi(p_s/2) = p_(s-n) where
pi(x) is the prime count function.

Anyways, let us pick the largest Ramanujan
prime < next unknown maximal prime gap prime, R_n < P_m where
P_m is the m-th record prime of p_(g+1) - p_g < p_(h+1) - p_h
for all g < h and G_m is the value of p_(h+1) - p_h at P_m.

Because all of the primes < next
unknown maximal prime gap prime, p < P_m, have prime gaps <= the
largest known maximal prime gap, all of the primes p are p_(s-n)
< p_(i-n) < p_s < P_m and have prime gaps g_(i-n) so that for
all s-n < i <= s are g_(i-n) <= G_(m-1).

Now each of the primes p_(i-n)
bounds the primes p_i by p_i < 2 p_(i-n), but p_(s-n) < p_(i-n)
< p_s, so p_(i-n+1) - p_(i-n) is < G_(m-1). This means that the
gaps are found to be 2 <= g_i < G_(m-1) unless it is a maximal
gap. And if it is a maximal gap then G_(m-1) < g_i = G_m < 2
* G_(m-1). Which is what we wanted to prove.

This might seem to be over
simplified, but it is what I have got for now.