Does $\sum^{\infty}_{n=1}\left(\frac{p_{n+1}}{p_n}-1\right)^2 $ converge? There appears to be a "prime constant" $\kappa$, generated from the sequence of primes:
$$\kappa = \sum^{\infty}_{n=1}\left(\frac{p_{n+1}}{p_n}-1\right)^2 \approx 1.653$$
Where $p_n$ is the nth prime.
However, how does one prove that such constant, does in fact, exist, that is, how does one prove that the above series converges?

See my answer to the similar problem here: Does this sum of prime numbers converge?
Note that $(\frac{p_{n+1}}{p_n}-1)^2 = \frac{(p_{n+1}-p_n)^2}{p_n^2}$.
Let $g_n=p_{n+1}-p_n$ be the sequence of prime gaps. Since $p_n\geq n$, the convergence would be proved if we have the convergence of
$$
\sum_{n=1}^{\infty} \frac{g_n^2}{n^2}.
$$
By this result of R. Heath-Brown: Here,
we have
$$
\sum_{n\leq x} g_n^2 \ll x^{\frac{23}{18}+\epsilon}.
$$
Applying the partial summation with $A(x)=\sum_{n\leq x}g_n^2$ and $f(x)=x^{-2}$, we obtain
$$
\sum_{n\leq x}\frac{g_n^2}{n^2}=\int_{1-}^x f(x)dA(x)
$$
$$
=A(x)f(x)-\int_{1-}^x A(t)f'(t)dt. \ \ (1)
$$
Since $3-(23/18) >1$, we obtain the convergence of (1). Hence, the desired sum converges by the comparison test.
