Does $\sum_{n=1}^\infty \frac{n}{{p_n}^{g_n}}$ converge? Let $p_n$ be the $n^\text{th}$ prime and $g_n$ be the $n^\text{th}$ prime gap ($p_{n+1} - p_n$). Calculating the serie up to $n = 10^6$ it seems that
$$\sum_{n=1}^\infty \frac{n}{{p_n}^{g_n}} < 1$$
Could this be true and is there a way to prove it?
(This is a follow up question to this, although perhaps a harder nut to crack.)
 A: Note that $g_n\ge 2$ except at the beginning. Also, by Rosser's Theorem we have $p_n\gt n\log n$. Then the result follows by Comparison with the convergent series $\sum_2^\infty \frac{1}{n\log^2 n}$.  (One needs much less than Rosser's Theorem, the Prime Number Theorem is enough.)
A: We have $g_n\geq 2$ so $\frac{1}{p_n^{g_n}} \leq \frac{1}{p_n^2}$. Further we have $p_n> n\log n$ for $n\geq 1$ by the prime number theorem (or Rosser's theorem to be precice). This implies that
$$S \leq \sum_{n=1}^{k}\frac{n}{p_n^{g_n}} + \sum_{n=k}^\infty\frac{1}{n(\log n)^2} \leq \sum_{n=1}^{k}\frac{n}{p_n^{g_n}}  + \frac{1}{\log(k)}$$
since $ \sum_{n=k}^\infty\frac{1}{n(\log n)^2} \leq \int_k^\infty \frac{dx}{x\log^2(x)} = \frac{1}{\log(k)}$. Computing the first sum above numerically we get
$$S \leq 1.355$$
for $k=10$. To push the bound below $1$ seems to require more precise estimates.
A: By Chebyshev's theorem (the weak version of the PNT) it follows that $p_n \geq C n \log n$ for some constant $C$ close to $1$, for any $n$ big enough. Since $g_n\geq 2$ for any $n\geq 2$ and the series 
$$\sum_{n\geq 2}\frac{1}{n\log^2(n)}$$
is convergent by Cauchy's condensation test, your series is convergent, too. 
Is the exact value of the original series really relevant for some purpose?
