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 divergent series related to the number of divisors of of p-1
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.


