Showing sum of reciprocals of primes less than $2^{100}$ is less than $8$ The question is: 
Let $P = {2, 3, 5, 7, 11,...}$ denote the set of all primes less than $2^{100}$. Show that 
$$\sum_{p\in P} \frac{1}{p} < 8$$
I've looked through some articles about prime numbers to approach this, for example I tried using the fact that the average gap between consecutive prime numbers among the first $N$ integers is roughly $\log(N)$. However, I'm still not quite sure how to approach this. Any help would be appreciated.
 A: We have
$$
\frac x{\log x-1}\lt\pi(x)\lt\frac x{\log x-1.1}
$$
for $x\ge60184$ (inequality $(6.6)$ in Estimates of Some Functions Over Primes without R.H. by Pierre Dusart), and thus
$$
\pi\left(2^k\right)-\pi\left(2^{k-1}\right)\lt\frac{2^k}{k\log2-1.1}-\frac{2^{k-1}}{(k-1)\log2-1}\;,
$$
so the sum of the reciprocals of the primes between $2^{k-1}$ and $2^k$ is bounded above by
$$
\frac2{k\log2-1.1}-\frac1{(k-1)\log2-1}\lt\frac2{(k-2)\log2}-\frac1{(k-2)\log2}=\frac1{(k-2)\log2}\;.
$$
Summing this for $k=17$ to $100$ yields
$$
\sum_{k=17}^{100}\frac1{(k-2)\log2}=\frac{H_{98}-H_{14}}{\log2}\lesssim2.77\;,
$$
where $H_n$ is the $n$-th harmonic number. We can readily sum the reciprocals of the remaining primes from $1$ to $2^{16}$; here’s Sage code for that:
j = var('j')
P=Primes()
sum ([1/P.unrank(j) for j in [0..6542]]).numerical_approx()

The result is
$$
\sum_{p\lt2^{16}}\frac1p\lesssim2.67\;,
$$
so in total the reciprocals of the primes up to $2^{100}$ sum to less than $2.77+2.67=5.44$.
