Sharpest bound on the zero free region of $\zeta^{\prime}$? I'm interested in calculating all of the zeroes of the first derivative of the Riemann $\zeta$ function up to an arbitrary height. I know that (on the RH), all of these zeroes will have real part $\ge \frac{1}{2}$. I am curious if there are equally strong upper bounds.
According to Titchmarsh, there is a constant c ($2<c<3$) such that every zero of $\zeta$'(s) has real part less than c. A. L. Skorokhodov, Pade approximants and numerical analysis of the Riemann zeta function , Zh. Vychisl. Mat. Mat. Fiz., 43, No. 9, 1330 – 1352, 2003 shows that $c\leq 2.93938$.
Has there been any improvement on this?
 A: If $s=\sigma+it$ with $\sigma >1$ then note that 
$$ 
\Big| \frac{\zeta^{\prime}}{\zeta}(s) \Big| =\Big| \sum_{p} \frac{\log p}{p^s-1} \Big| 
\ge \frac{\log 2}{|2^s-1|} - \sum_{p\ge 3} \frac{\log p}{|p^{s}-1|} 
\ge \frac{\log 2}{2^{\sigma}+1} - \sum_{p\ge 3} \frac{\log p}{p^{\sigma}-1}.
$$ 
Thus 
$$ 
\Big|\frac{\zeta^{\prime}}{\zeta}(s) \Big| \ge \frac{\log 2}{2^{\sigma}+1} + \frac{\log 2}{2^{\sigma}-1} + \frac{\zeta^{\prime}}{\zeta}(\sigma) = 
\frac{2^{\sigma+1}\log 2}{4^{\sigma}-1} + \frac{\zeta^{\prime}}{\zeta}(\sigma).
$$ 
A calculation shows that the RHS above equals zero for $\sigma = 2.813\ldots$, and is strictly positive for larger values of $\sigma$.  Therefore any zero of $\zeta^{\prime}(s)$ has real part at most $2.813\ldots$.   
Moreover, the inequalities above are essentially tight if $t$ is chosen so that $2^{it} \approx -1$ but $p^{it} \approx 1$ for many small primes $3\le p\le z$ (say, with $z$ large).  Choosing such $t$, and using a Rouche's theorem argument one can show that there will be zeros of $\zeta^{\prime}$ with real part arbitrarily close to $2.813\ldots$.  
Interestingly, this argument is essentially given in Titchmarsh's book, see Theorem 11.5 A of The theory of the Riemann zeta-function.  The more recent result of Skorokhodov is unfortunately a worse argument than Titchmarsh.  The reason is that Skorokhodov argues with $\zeta^{\prime}(s) =- \sum_{n=2}^{\infty} (\log n)/n^s$, but the values of $n^{it}$ are not independent; arguing with primes is better as the values $p^{it}$ do behave independently.  
