I'm following this paper: http://mathdl.maa.org/images/upload_library/22/Chauvenet/Zagier.pdf
Define $\Phi(s) = \displaystyle\sum_p \frac{\log p}{p^s}$.
By taking a logarithm and differentiating the Euler product formula $\zeta(s) = \displaystyle\prod_p \frac{1}{1-p^{-s}}$, we derive
-\frac{\zeta'(s)}{\zeta(s)} = \Phi(s) + \displaystyle\sum_p \frac{\log p}{p^s(p^s-1)}.
Assume that $s_0 = 1 + ia$ is a zero of $\zeta$ of order $\mu$. Then we can write $\zeta(s) = (s-s_0)^{\mu} g(s)$ for some function $g(s)$ that is analytic and nonzero at $s_0$.
So \zeta'(s) = \mu(s-s_0)^{\mu -1} g(s) + g'(s) (s-s_0)^{\mu} by the product rule.
With a little algebra you can derive
-\frac{\zeta'(s)}{\zeta(s)} = -\frac{\mu}{s-s_0} - \frac{g'(s)}{g(s)}.
This yields the formula
\Phi(s) = -\frac{\mu}{s-s_0} - \frac{g'(s)}{g(s)} - \displaystyle\sum_p \frac{\log p}{p^s(p^s-1)}.
Define F(s) = -\frac{g'(s)}{g(s)} - \displaystyle\sum_p \frac{\log(p)}{p^s(p^s-1)}. Now we can show that
$\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon + ia) = \displaystyle\lim_{\epsilon \searrow 0} \epsilon F(1+\epsilon+ia) - \mu = - \mu.$
through direct computation.
Now, they make a claim in the paper that "because $s=1$ is a simple pole of $\zeta$ of residue $1$", we have $\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon) = 1$. This is what I don't understand. Doing a similar process to what I did above gives me
\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon) = \displaystyle\lim_{\epsilon \searrow 0} -\frac{\epsilon \mu}{\epsilon - ia} - \epsilon \frac{g'(1+\epsilon)}{g(1+\epsilon)} - \epsilon \displaystyle\sum_p \frac{\log p}{p^{1+\epsilon}(p^{1+\epsilon}-1)}
And I don't see any way I can get a $1$ as the result, since all terms should converge to zero. The second and third terms will go to zero because they did before and the first term will go to zero by simple limit laws.
I conclude I have made a mistake someplace or I am missing something obvious.