Olympiad problem on the prime numbers

Let $P={2,3,5,7,11,...}$ denote the set of all prime numbers less than ${ 2 }^{ 100}$.

Prove that $\sum _{ p\in P }^{ }{ \frac { 1 }{ p } } < 8$.

I don't understand how to progress in the problem. Any help would appreciated. Thank you.

Rosser and Schoenfeld in Approximate formulas for some functions of prime numbers give an explicit form of Mertens' second theorem, an upper bound valid for all $x>1$: $$ \sum_{p\le x} \frac1p < \log \log x + B + \frac1{\log^2x} $$ where $B \approx 0.26149\cdots$. This gives $$ \sum_{p\le 2^{100}} \frac1p < 4.51 < 8 $$ However, this argument is most probably not in the spirit of the question.

The problem can be easily solved if you can use the following bound on prime counting function $\pi(n)$ (taken from here): $$\pi(n) / \frac{n}{\ln{n}} \le C = 1.25506$$

Rewrite you sum via $\pi(n)$: $$ \sum_{p \in P} \frac{1}{p} = \sum_{k=1}^{2^{100}}\frac{\pi(k) - \pi(k-1)}{k} $$

Now use summation by parts, with $k$ starting from $a > 1$: $$ \sum_{k=a}^{2^{100}}\frac{\pi(k) - \pi(k-1)}{k} = \left[ \frac{\pi(2^{100})}{2^{100} + 1} - \frac{\pi(a-1)}{a} \right] - \sum_{k=a}^{2^{100}}\pi(k)\left(\frac{1}{k+1} - \frac{1}{k}\right) $$ Forget about the part in brackets, it should be easy to bound efficiently. The sum in the main part can be bounded: $$ -\sum_{k=a}^{2^{100}}\pi(k)\left(\frac{1}{k+1} - \frac{1}{k}\right) = \sum_{k=a}^{2^{100}}\frac{\pi(k)}{k(k+1)} \le \sum_{k=a}^{2^{100}}\frac{C \; k}{k(k+1)\ln{k}} \le C \sum_{k=a}^{2^{100}}\frac{1}{k\ln{k}} $$ We can bound this sum via integration. Define function $f(x) = (x \ln x)^{-1}$. It monotonically decreases for $x > 1$, so $(k \ln k)^{-1} \le \int_{k-1}^{k} f(x) dx$. Hence: $$ \sum_{k=a}^{2^{100}}\frac{1}{k\ln{k}} \le \int\limits_{a-1}^{2^{100}-1} \frac{dx}{x \ln x} = \ln \ln x \bigg\rvert_{a-1}^{2^{100}-1} \le \ln \ln 2^{100} = 4.23865\ldots $$ Multiplying this value by $C$ given above, we get a bound $5.32 < 8$. However, we also have to add the part in the brackets and the original sum for $k < a$.