Fraction of prime number between consecutive powers of two A $n$-bit integer is an integer $x$ such that $2^{n-1} \le x < 2^n$. In [1] it is claimed without proof that a corollary of the prime number theorem is that:

For any $n > 1$, the fraction of $n$-bit integers that are prime is at least $\frac{1}{3n}$.

Can somebody point me to a proof of this?
It is (perhaps) worth noticing that a direct application of the prime number theorem allows to prove an asymptotic version of the above (with a slightly better constant).
$$
\begin{align*}
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \pi(2^n-1) - \pi(2^{n-1})\bigg) &= 
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg(  \pi(2^n) - \pi(2^{n-1}) \bigg) \\&=
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}}  \bigg( \frac{2^n}{n \ln 2}  - \frac{2^{n-1}}{(n-1)\log2} \bigg)\\ & =
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}}  \bigg( \frac{2^n}{2n \ln 2 } \cdot\frac{n-2}{n-1}   \bigg) = 1.
\end{align*}
$$
Therefore $\frac{\pi(2^n-1) - \pi(2^{n-1})}{2^{n-1}} \sim \frac{1}{n \ln 2}$, and for every $\varepsilon >0$ there is some $n_0$ such that, for all $n \ge n_0$, the fraction of $n$-bit integers that are prime is at least $\frac{1-\varepsilon}{n \ln 2}$.
A not-so-elegant way to prove the the claim would be that of choosing $\varepsilon =1-\frac{\ln 2}{3} \approx 0.7689$ and finding an upper bound on $n_0$ that lies in the range for which $\pi(\cdot)$ has been computed. Are such upper bounds known?
[1] Jonathan Katz, Yehuda Lindell. Introduction to Modern Cryptography (3rd edition).  CRC Press. ISBN 9781351133012.

According to
the classic article
"Explicit Bounds for Some Functions of Prime Numbers",
Barkley Rosser
American Journal of Mathematics
Vol. 63, No. 1 (Jan., 1941), pp. 211-232 (22 pages)
for $x \ge 55$,
$\dfrac{x}{\log x+2}
\lt \pi(x)
\lt \dfrac{ x}{\log x-4}
$.
Therefore,
putting $x=2^n$,
$\dfrac{ 2^n}{n\log 2+2}
\lt \pi(2^n)
\lt \dfrac{ 2^n}{n\log 2-4}
$.
Therefore
$\dfrac{ 2^n}{n\log 2+2}-\dfrac{ 2^{n-1}}{(n-1)\log 2-4}
\lt \pi(2^n)-\pi(2^{n-1})
\lt \dfrac{ 2^n}{n\log 2-4}-\dfrac{ 2^{n-1}}{(n-1)\log 2+2}
$.
Those differences are about
(more explicitness later)
$\begin{array}\\
D
&=\dfrac{ 2^n}{n\log 2}-\dfrac{ 2^{n-1}}{(n-1)\log 2}\\
&=\dfrac{ 2^n}{n\log 2}\left(1-\dfrac{\frac12}{1-\frac1{2n}}\right)\\
&=\dfrac{ 2^n}{n\log 2}\left(\dfrac{1-\frac1{2n}-\frac12}{1-\frac1{2n}}\right)\\
&=\dfrac{ 2^n}{n\log 2}\left(\dfrac{\frac12-\frac1{2n}}{1-\frac1{2n}}\right)\\
&=\dfrac{ 2^n}{2n\log 2}\left(\dfrac{1-\frac1{n}}{1-\frac1{2n}}\right)\\
&=\dfrac{ 2^n}{2n\log 2}\left(\dfrac{1-\frac1{2n}-\frac1{2n}}{1-\frac1{2n}}\right)\\
&=\dfrac{2^n}{2n\log 2}\left(1-\dfrac{\frac1{2n}}{1-\frac1{2n}}\right)\\
&=\dfrac{2^n}{2n\log 2}\left(1-\dfrac{1}{2n-1}\right)\\
&\approx\dfrac{ 2^n}{2n\log 2}\\
\end{array}
$
For the lower bound,
$\begin{array}\\
L(n)
&=\dfrac{ 2^n}{n\log 2+2}-\dfrac{ 2^{n-1}}{(n-1)\log 2-4}\\
&=\dfrac{ 2^n((n-1)\log 2-4)-2^{n-1}(n\log 2+2)}{(n\log 2+2)((n-1)\log 2-4)}\\
&=\dfrac{ 2^{n-1}(2(n-1)\log 2-8)-(n\log 2+2))}{n(n-1)\log^22-4n\log 2+2(n-1)\log 2-8}\\
&=\dfrac{2^{n-1}(n \log 2-2\log 2-10)}{n(n-1)\log^22-2n\log 2+2\log 2-8}\\
\end{array}
$
I will leave it at this for now,
and a similar expression
can be derived for the
upper bound.
