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.

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).