Conjecture: the sequence of sums of all consecutive primes contains an infinite number of primes

As Will Jagy points out, a proof is probably out of reach, but the statement is probably true.

One can give a heuristic argument. The $n$-th prime is roughtly $n\log n$, so $S_n$ is roughly

$$S_n \approx \sum_{i=1}^n i \log i \approx \frac{n^2(2\log n - 1)}{4}.$$

The density of primes around $x$ is approximately $1/\log x$, so the probability that $S_n$ is prime is roughly

$$1/\log\left(\frac{n^2(2\log n - 1)}{4}\right) \approx \frac{1}{2\log n + \log(2\log n-1) - \log 4} \approx \frac{1}{2\log n}.$$

So the expected number of primes among $S_1, \dots, S_n$ is roughly

$$\sum_{i=1}^n \frac{1}{2\log i} \approx \frac{\text{li}(n)}{2},$$

which goes to $\infty$ as $n\to \infty$.

(I just saw that Greg Martin came to the same conclusion in the comments.)

The recent paper "Curious conjectures on the distribution of primes among the sums of the first $2n$ primes" by Romeo Meštrović at https://arxiv.org/abs/1804.04198 states your conjecture, along with certain details like the estimate of the number of primes, in section 3. This paper's other sections examine the distribution of primes, using heuristic, computational and analytic arguments that they are similar to the distribution of primes among the positive integers.