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?

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

1 Answer 1


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.

  • $\begingroup$ I don't understand yet the proof, but I'm really thankful to know that it is convergent! $\endgroup$
    – ordptt
    Feb 13, 2021 at 21:56

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