TITLE: Does $\sum^{\infty}_{n=1}\left(\frac{p_{n+1}}{p_n}-1\right)^2 $ converge?
QUESTION [10 upvotes]: 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?

REPLY [4 votes]: 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.