A divergent series related to the number of divisors of of p-1 Let $d(n)$ denote the number of divisors of $n$. Is it known that the series
$$\sum_{p \text{ prime}} \frac{1}{d(p-1)}$$
diverges?
This would follow immediately from the Sophie Germain Conjecture. Indeed, if there are infinitely many primes of the form $2p+1$ ($p$ a prime), then infinitely many terms of the series are equal to $1/4$, so the series doesn't even satisfy the most basic requirement for convergence! So, surely there must be a direct proof?
 A: Since every divisor $k$ of $p-1$ either satisfies $k\le \sqrt{p-1}$ or $\frac{p-1}{k}\le \sqrt{p-1}$, we have $d(p-1) \le 2\sqrt{p-1}$. If we let $p_n$ denote the $n$th prime, then since $p_n-1 \le n^2$ for all $n$ (an easy consequence of the prime number theroem...), we have
$\sum_{n=1}^N \frac{1}{d(p_n-1)} \ge \sum_{n=1}^N\frac{1}{2n} \ge \frac{\ln N}{2}$,
so the partial sums diverge.
Obviously one can get much better bounds than these.
A: Answering my own question, because I totally overlooked the following ridiculous idea:
Obviously $d(n)\leq n$ for every $n$. Thus $d(p-1)\leq p-1 < p$, so $1/d(p-1) > 1/p$ and the divergence follows from the divergence of $\sum 1/p$ (if one is willing to assume that).
A: I did a heuristic analysis to study how this sum is growing. I calculated $d(p_n-1)$ and the sum and plotted the curve of the sum vs $n$. I obtained a very smooth curve which looked like a cumulative distribution curve. Next I did curve fitting to model this curve where the sum is the dependent variable and $n$ is the independent variable. The boundary condition I imposed was that the sum should be divergent. The following model was found to best fit the desired sum. 
$$
\sum_{n=1}^{x}\frac{1}{d(p_n - 1)} \sim e^{a + b/x + c\ln\ln x}
$$
where $a,b$ and $c$ are suitable constants. 
The true asymptotic formula could be different from the above but I believe this can give some hints in the direction of the true asymptotics.
