How I can calculate $ \lim_{n→∞}\frac{\ln(2^{p_{n+1}}-1)}{\ln(2^{p_{n}}-1)} $? Let $\left\{\, p_{n}\,\right\}$ be the sequence of consecutive primes.
$$\mbox{How I can calculate}\quad
\lim_{n\ \to\ \infty}{\ln\left(\, 2^{p_{n + 1}} - 1\,\right)\over
                       \ln\left(\, 2^{p_{n}} - 1\,\right)}\ {\large ?}
$$
I've tried to use the fact that
$\displaystyle{\lim_{n\ \to\ \infty}p_{n} = +\infty}$ but I cannot arrive at any result.
 A: $$\lim_{n\to\infty}\frac{\ln(2^{p_{n+1}}-1)}{\ln(2^{p_{n}}-1)}=\lim_{n\to\infty}\frac{\ln(2^{p_{n+1}})}{\ln(2^{p_{n}})}=\lim_{n\to\infty}\frac{p_{n+1}}{p_n}$$
In Analytic Number Theory, of Tom M. Apostol, p. 84 there is an useful estimate:
$$\frac16n\ln n<p_n<12\left(n\ln n+n\ln(12/e)\right)$$
With this estimate you can easily see that $$\lim_{n\to\infty}\frac{p_{n+1}}{p_n}=1$$
A: HINT. $$p_n \sim n\ln(n),$$
(see Prime number theorem).
Then
$$\frac{\ln(2^{p_{n+1}}-1)}{\ln(2^{p_{n}}-1)}\sim \frac{\ln(2^{p_{n+1}})}{\ln(2^{p_{n}})}=\frac{p_{n+1}}{p_n}\rightarrow 1$$
A: Following the hint in the comments: 
$$\lim_{n→∞}\frac{\ln(2^{p_{n+1}}-1)}{\ln(2^{p_{n}}-1)}$$
$$= \lim  \frac{p_1\log 2 + \log (1 - 2^{-p_n})}{p_{n+1}\log 2+ \log (1 - 2^{-p_2})}$$
As $ n \to \infty, \log(1 - 2^{-p_n})\to \log 1 = 0. $ Then we have
$$ =\lim \frac{p_n\log 2}{p_{n+1}\log 2} $$
and we should not neglect to show that $\frac{p_n}{p_{n+1}}\sim 1.$
$$\lim \frac{p_n}{p_{n+1}} = \lim \frac{p_n}{n \log n}\cdot\frac{(n+1)\log (n+1)}{p_{n+1}}\cdot \frac{n\log n}{(n+1)\log (n+1)}= \lim \frac{n\log n}{(n+1)\log (n+1)}$$ the last equality via the PNT. 
Since the last is indeterminate ($\frac{\infty}{\infty}$) we can use L'Hopital:
$$\lim_{n\to \infty}\frac{n\log n}{(n+1)\log (n+1)} = \lim \frac{1\cdot \log n+n\cdot\frac{1}{n}}{1\cdot\log(n+1)+(n+1)\frac{1}{n+1}}$$
$$= \lim \frac{\log n + 1}{\log (n+1)+1} $$
We apply L'Hopital again--
$$\lim \frac{1/n}{1/(n+1)} = \lim \frac{n+1}{n}=\lim 1+\frac{1}{n} = 1 $$
