This is a re-post with my (corrected) earlier typo in red as a reminder. The question is about
$$(A)\hspace{10mm}\lim_{n \to \infty} \sum_{k=\color{red}2}^{n} \frac{p(k +1) - p(k)}{p(k)^2} = \frac{1}{2}.$$
in which $p(k)$ is the $k^{th}$ prime.
In the original post (whose sum began at $k=1$) I explained why I thought the sum on the left converged and it seemed obvious enough to the person who answered so I'll omit that here.
My question is whether there is a possibly unproven idea that suggests this is correct?
There are many interesting results about the numerator and some well-known conjectures but nothing that seems to give this.
For n = 500, 5000, 50000, I get Sum $\approx$ 0.49069, 0.49095, 0.49097, respectively.
This time I am including some Mathematica code and a picture:
S = Table[ Sum[(Prime[i + 1] - Prime[i])/Prime[i]^2, {i, 2, j}], {j, 2, 500}];
Thanks for any insights.
Note that $\dfrac{p(k+1)}{p(k)} < 2$ for all $k$. Thus, $\displaystyle\sum_{k = m}^{\infty}\dfrac{p(k+1)-p(k)}{p(k)^2} = \sum_{k = m}^{\infty}\dfrac{p(k+1)-p(k)}{p(k+1)^2} \cdot \dfrac{p(k+1)^2}{p(k)^2}$ $\displaystyle \le \sum_{k = m}^{\infty}\dfrac{p(k+1)-p(k)}{p(k+1)^2} \cdot 2^2 \le 4\sum_{k = m}^{\infty}\int_{p(k)}^{p(k+1)}\dfrac{\,dx}{x^2} = 4\int_{p(m)}^{\infty}\dfrac{\,dx}{x^2} = \dfrac{4}{p(m)} $.
Thus, $\displaystyle\sum_{k = 2}^{\infty}\dfrac{p(k+1)-p(k)}{p(k)^2} \le \dfrac{4}{p(m)} + \sum_{k = 2}^{m-1}\dfrac{p(k+1)-p(k)}{p(k)^2} $ for any integer $m$,
Taking $m = 100$ yields $\displaystyle\sum_{k = 2}^{\infty}\dfrac{p(k+1)-p(k)}{p(k)^2} \le 0.496504 < \dfrac{1}{2}$.
Hence, the sum does not converge to $\dfrac{1}{2}$.