I ask if the series $$\sum_{k=1}^{\infty}\frac{p_{k+1}-p_k}{(p_{k+1}+p_k)^\alpha}$$ where $p_k$ stands for the prime of index $k$, has the same properties of convergence of the series $$\sum_{k=1}^{\infty}\frac{1}{k^\alpha}$$ that is convergent for all $\alpha \gt 1$ and divergent for all $\alpha \le 1$.

In the case $\alpha = 1$, I conjecture the following asymptotic behavior of the sum of the series $$\sum_{k=1}^{n}\frac{p_{k+1}-p_k}{p_{k+1}+p_k}\sim \gamma \log n$$ while in the case $\alpha = 2$ the series seems to converge to the value $0.1200307...$


1 Answer 1


Let $p$ denote a prime and $p'$ denote the next prime. Let $x>1$ be a large parameter.

By the positivity of $p'-p$ and the fact that $p'\sim p$, $$\sum_{x\leq p<2x}\frac{p'-p}{(p'+p)^\alpha}\asymp x^{-\alpha}\sum_{x\leq p<2x}(p'-p)\asymp x^{1-\alpha}.$$ Hence, applying a dyadic decomposition, it follows that $$\sum_{k=1}^{\infty}\frac{p_{k+1}-p_k}{(p_{k+1}+p_k)^\alpha}$$ converges for $\alpha>1$, but diverges for $\alpha\leq 1$.


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