This question was also asked on MSE. Does there exist an asymptotic estimate for the following sum over primes $$ \sum_{p\leq x} \frac{\tau(p-1)}{p}\;, $$ where $\tau(n)=\sum_{d|n}1$ is the divisor function?
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5$\begingroup$ math.stackexchange.com/questions/928906/… $\endgroup$– Will JagyCommented Sep 12, 2014 at 17:40
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8$\begingroup$ Please do not ask questions simultaneously on MathOverflow and Math.StackExchange. It leads to duplication of effort and is frowned upon by both communities. Please pick one site and wait at least a few days for an answer before reposting the question elsewhere. $\endgroup$– Ricardo AndradeCommented Sep 12, 2014 at 18:29
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For this kind of things it's always a good idea to check in the two volumes of the Handbook of Number Theory of J. Sandor, D. S. Mitrinovic and B. Crstici. Here I found the formula:
$$(\star)\quad \sum_{p \leq x} \tau(p - 1) = \frac{315 \,\zeta(3)}{2 \pi^4} \cdot x + O\!\left(\frac{x}{(\log x)^\alpha}\right), $$ as $x \to +\infty$, for any $\alpha \in \;]0,1[$. Yu. V. Linnik. New versions and new uses of the dispersion method in binary additive problems (Russian). Dokl. Akad. Nauk SSSR 137 (1961), 1299–1302.
Using ($\star$) and partial summation (Tenenbaum - Introduction to Analytic and Probabilistic Number Theory, Chapter I.0, Theorem 1 with $a_n := \tau(n-1)$ if $n$ is prime, $a_n := 0$ is $n$ is composite and $b(t) := 1/t$) you should be able to get an asymptotic formula for $\sum_{p \leq x} \tau(p-1)/p$.