Are there any bounds on residues of $1/\zeta$ in roots of $\zeta$ in critical strip, which may use RH, but do not use the conjecture on simplicity of roots or something similar? I did not find such resuts in Titchmarsh, but I could miss something. Thanks!



Residues of $1/\zeta$
This is actually a very difficult problem, and currently most results are highly conjectural. It essentially comes down to finding useful bounds on discrete moments of the Riemann zeta function of the form
$$J\_k(T) = \sum\_{0 < \Im(\rho) < T}{|\zeta'(\rho)|^{2k}},$$
as one can then choose the correct value of $k$ and apply partial summation.


For positive $k$, recent results of [Milinovich](http://arxiv.org/abs/0806.0786) and [Milinovich and Ng](http://arxiv.org/abs/0706.2321) show under the Riemann Hypothesis that
$$T(\log T)^{(k+1)^2} \ll J\_k(T) \ll T(\log T)^{(k+1)^2 + O(1/\log \log \log T)};$$
and slightly stronger results are known for $k = 0,1,2$ (with the latter two being under the Riemann Hypothesis).


For negative $k$, the situation is much more difficult. A conjecture of Gonek-Hejhal suggests that for all $k > -3/2$,
$$J\_k(T) \asymp T(\log T)^{(k+1)^2}$$
and this has been refined significantly by [Hughes, Keating, and O'Connell](http://www.jstor.org/stable/2665448). But it seems out of reach to prove anything significantly useful in this area; the best result so far has been by [Gonek](http://journals.cambridge.org/production/action/cjoGetFulltext?fulltextid=7010328), who proved that $J\_{-1}(T) \gg T$ assuming the Riemann Hypothesis and the simplicity of the zeroes of $\zeta(s)$. But this isn't useful for most applications, where an upper bound is needed. I believe it is possible to show $J\_{-1} \ll T^{2+\varepsilon}$ under the Riemann Hypothesis and the simplicity of the zeroes of $\zeta(s)$, though I don't have a reference for this. Also it is quite possible that this result also holds if we replace $|\zeta'(\rho)|^{-2}$ by $\left|\mathrm{Res}\_{s = \rho} \zeta(\rho)^{-1}\right|^2$, though again I don't know of a reference for this.


