Calculating $\dfrac{d^{m-1}}{ds^{m-1}} \Big(\dfrac{\zeta'(s)}{\zeta(s)} \Big)$

Question

Eq. 1.4 in Levinson's paper states $$\dfrac{(-1)^m}{m!} \dfrac{d^{m-1}}{ds^{m-1}} \Big(\dfrac{\zeta'(s)}{\zeta(s)} \Big) = \dfrac{1}{(s-1)^m} - \sum_{n=0}^{\infty} \dfrac{1}{(s+2n)^m} - \sum_{\rho} \dfrac{1}{(s-\rho)^m}.$$ By Eq. 2.12.7 in Titchmarsh's book The Theory of the Riemann Zeta-Function, $$\dfrac{\zeta'(s)}{\zeta(s)} = b - \dfrac{1}{(s-1)} -\dfrac{1}{2} \dfrac{\Gamma'(\frac12 s +1)}{\Gamma(\frac12 s +1)} + \sum_{\rho} \Big( \dfrac{1}{s-\rho} + \dfrac{1}{\rho} \Big).$$ And at page 235 of Whittaker & Watson's book A Course of Modern Analysis, we see $$\dfrac{d^2}{dz^2} \log \Gamma(z+1) = \sum_{n=1}^{\infty} \dfrac{1}{(z+n)^2}.$$ I calculated for many times and still I get a different series than $\sum_{n=0}^{\infty} \dfrac{1}{(s+2n)^m}$ stated in Levinson's paper. For example, for $m=3$ I got $$\dfrac{(-1)^3}{3!} \dfrac{d^2}{ds^2} \Big( -\dfrac{1}{2} \dfrac{\Gamma'(\frac12 s +1)}{\Gamma(\frac12 s +1)} \Big) = \dfrac{(-1)^3}{3!} \dfrac{1}{2^3} \sum_{n=0}^{\infty} \dfrac{2}{(\frac12 s +n)^3},$$ which differs by a factor of $2$, and also $n$ starts from $1$ not $0$. Is the paper wrong or me?