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 $0.1200307...$



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


