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The prime gaps grow logarithmically. Now, suppose I create a list of $N$ primes. For example $N = 10$ or $[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]$ then

$$\text{total~sum} = 129$$ $$2^N = 1024$$

Question

Will the $total~sum$ of the list always be less than $2^N$?

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A very elementary way to prove it is to observer that starting with $p_4=7$ $$p_n<2^{n-1} \tag{1}$$ which can be shown by induction and using Bertrand's postulate $$p_{n+1}< 2p_n<2^{n}$$ Then $\forall N\geq 6$ we have $$\sum\limits_{k=1}^N p_k =\sum\limits_{k=1}^5 p_k + \sum\limits_{k=6}^N p_k =28+\sum\limits_{k=6}^N p_k< 28+\sum\limits_{k=6}^N 2^{k-1}<\\ 31+\sum\limits_{k=6}^N 2^{k-1}=\sum\limits_{k=1}^N 2^{k-1}=\frac{2^N-1}{2-1}=2^N-1<2^N$$ Of course, $N=5$ can be tested manually.

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