Why does this particular ratio of prime numbers seems to converge to 3? I noticed this interesting property of prime numbers, and I'd like to know if it has an explanation/proof/disproof.
Define $p(n)$ to be the $n$'th prime number.
Define the following sequence: 
 $$\Sigma(n) =
\begin{cases}
p(1),  & \text{if $n=1$} \\
\Sigma(n-1)+p(n), & \text{if $n>1$ and  $\Sigma(n-1)-p(n)<0$} \\
\Sigma(n-1)-p(n), & \text{otherwise}
\end{cases}$$
The first few elements of the sequence are:  $2,5,0,7,18,5,22,3,26,55$.
Now, in $\Sigma(n)$ lets look at all indexes $n$'s such that $\Sigma(n-1)<\Sigma(n)<\Sigma(n+1)$. These indexes also form a sequence, which I'll denote by $a(k)$. Here are its first elements: $4,9,22,57,146,367,946,2507$.
So, what I noticed is that these two limits seem to hold: 
$$\lim_{k\to \infty}\frac{p(a(k+1))}{p(a(k))} = 3$$
$$\lim_{k\to \infty}\frac{\Sigma(a(k+1))}{\Sigma(a(k))} = 3$$
Here is a graph of the former of these ratios:

Of course, these are only empirical findings. Do you have other reasons to believe that they are true?
 A: OK, I got the idea of where the universal constant 3 comes from, but not the proof for $p_n$.
Let $d_n$ be a strictly increasing sequence of natural numbers, such that the differences $d_{n+1}-d_n$ are asymptotically "small" compared to $d_n$ themselves. Let's form a series of partial sums of $\pm d_n$ according to the same rules: minus unless this would drive the sum into the negatives, otherwise plus. (On a side note, I don't like the idea of non-standard usage of $\Sigma$, but also I don't like to introduce extra symbols unless absolutely necessary, so I will just avoid mentioning these partial sums other than verbally). Now what does it mean when a partial sum has two growth steps in a row?
Most of the time, the signs just follow a dull alternating sequence: +, -, +, -, +, -... Every once in a while, the partial sum drops so low that the next  (+, -) would drive it below 0 (which means that the sum was really low, less than $d_{n+1}-d_n$, which itself is $o(d_n)$, like I said before). Then we throw in an extra plus (so the sum which was basically $0$ becomes basically $d_n$), and then continue with the dull +, -, +, -, +, -... BTW, that's when $n$ becomes the next term in the $a_k$ series. Now, each (+, -) brings the sum a little lower, because $d_i-d_{i+1}<0$. How long it might be before these bites deplete our stock and the sum is driven close to 0 again? If we would add every difference, they would all telescope nicely, and we'd have $d_{next}-d_n\approx d_n$, or $d_{next}\approx 2d_n$. But we use just every other difference. Now, if the sequence grows relatively "uniformly" (which is easy to prove for $d_n=n$ or $n^2$, but not easy at all for $d_n=p_n$), then the sum of every other difference is about half of sum of all differences, so it boils down to ${1\over2}(d_{next}-d_n)\approx d_n$, or $d_{next}\approx \color{red}{\bf3}d_n$.
A: Say for some $n$:
$$\Sigma(n-1)<\Sigma(n)<\Sigma(n+1)$$
Then: 
$$\Sigma(n+1)=\Sigma(n-1)+p(n)+p(n+1)\geq p(n)+p(n+1)$$
Suppose $n>25$. Now $p(n)>\dfrac{5}{6}p(n+1)$ and $p(n+2)<\frac65p(n+1)<\frac{36}{25}p(n)$. Hence:
$$p(n)+p(n+1)>\Big(\frac56+1\Big)p(n)>\frac{36}{25}p(n)>p(n+2)$$
So $\Sigma(n+1)>p(n+2).$ Therefore $\Sigma(n)$ cannot increase more than $2$ times in a row for $n>25$.

Say for some $n$:
$$\Sigma(n-1)>\Sigma(n)>\Sigma(n+1)$$
Then:
$$\Sigma(n-1)\geq p(n)+p(n+1)>p(n-1)+p(n)$$
Therefore $\Sigma(n-2)>p(n-1)$ and $\Sigma(n-1)=\Sigma(n-2)-p(n-1)$. So:
$$\Sigma(n-1)>\Sigma(n)>\Sigma(n+1)\implies \Sigma(n-2)>\Sigma(n-1)>\Sigma(n)$$
But since $\Sigma(n)$ decreases somewhere (For instance, $\Sigma(3)<\Sigma(2)$), there are not $n$ such that $\Sigma(n-1)>\Sigma(n)>\Sigma(n+1)$ and hence $\Sigma$ does never decrease twice in a row.

This confirms what Ivan Neretin says, namely that the pattern will be +,-+,-... and occasionally +,+,-, when $\Sigma(n-1)<p(n+1)-p(n)$.
