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.
 A: 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.
A: 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 $k < a$.
