# Convergence of series $\sum_{k=1}^{\infty}\frac{p_{k+1}-p_k}{(p_{k+1}+p_k)^\alpha}$

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 $$0.1200307...$$

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 $$\alpha\leq 1$$.