Is the sum of the first $n$ primes a prime infinitely many times? Define the sequence $P(n)=\sum_{i=1}^{n}p_i$, where $p_i$ is the $i$-th prime number.
I observed for some small $n$ that sometimes this sum evaluates to a prime number, for example $P(2)=2+3=5$ and $P(4)=2+3+5+7=17$ and $P(6)=2+3+5+7+11+13=41$.
So it is natural to ask:

Is it true that there is a sequence of natural numbers $\{n_i:i \in \mathbb N\}$ such that all numbers in the set $\{P(n_i):i \in \mathbb N\}$ are prime numbers?

 A: This is not an answer, but it intend to show that there seems to be an infinite number of primes of the form 
$\sum_{k=1}^n p_k$.
The x-axis in the diagram is $\,^2\!\log n$ and the y-axis is $\,^2\!\log N_n$, where $N_n$ is the number of primes of the form 
$\sum_{k=1}^m p_k$ where $m\le n$.

A: up to $n=1$ there are $1$ prime values, this is $100.000000$ percent
up to $n=10$ there are $4$ prime values, this is $40.000000$ percent
up to $n=100$ there are $10$ prime values, this is $10.000000$ percent
up to $n=1000$ there are $76$ prime values, this is $7.600000$ percent
A: Almost certainly yes, but I doubt very much that this can be proven at the current state of the art.
The first few primes that arise are 
$$2, 5, 17, 41, 197, 281, 7699, 8893, 22039, 24133, 25237, 28697, 32353, 37561, 38921, 43201, 44683, 55837, 61027, 66463, 70241, 86453$$
See OEIS sequence A013918.
The sum of the first $n$ primes is on the order of $n^2 \log n$, and heuristically a number of this size has probability on the order of $1/\log n$ of being prime.  Since $\sum_n 1/\log n = \infty$, there ought to be infinitely many.  But that's not a proof.
A: Not an answer, for the obvious reasons, stated by other people. But here is an attempt I am trying, unsuccessfully so far. There is this book "Problems in Real Analysis. Advanced Calculus on the Real Axis" by:


*

*Teodora-Liliana T. Radulescu

*Vicentiu D. Radulescu

*Titu Andreescu


And it contains an elegant short proof of the following:

Now, combining this with Legendre's conjecture, we obtain: 
$$\left | \sqrt{P(n)} - \sqrt{q} \right | < 1$$
where $q$ is some prime number within the same consecutive perfect squares as $P(n)$.
Obviously, not every $n$ will yield a possible prime to consider, because $P(n)$ is even for odd $n$. But, if by some magic density or Pigeonhole principle argument we could prove that $\left | \sqrt{P(n)} - \sqrt{q} \right |$ is close enough so that $\left | P(n) - q \right | < 1$, that would make it a nice proof, based on Legendre's conjecture of course.
