# 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?

• Have you considered the asymptotic behaviour of $p_n$ as given by the prime number theorem? And its relative error? Feb 13, 2021 at 19:41

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.