Evaluate or not to...

The Weierstrass form of the Gamma function is $\Gamma(z) = \left[z e^{\gamma z} \prod_{k=1}^\infty \left(1 + \frac zk\right) e^{-z/k} \right]^{-1}$ Taking its logarithmic differentation, $\dfrac{\Gamma'(z) }{\Gamma(z)} = \frac 1z + \gamma + \sum_{n=1}^\infty \left( \frac{1/n}{1 + z/n} - \frac1n \right).$ Thus, the first few terms of the Maclaurin series of $\Gamma(z)$ is $\Gamma(z) = \frac 1z + \gamma + z( \cdots)$ And because $\frac1{1+x} - 1 = -\sum_{k=2}^\infty (-1)^k x^{k-1}$ The following digamma series is a direct result, $\Psi(1+x) = -\gamma + \sum_{k=2}^\infty (-1)^k \zeta(k) x^{k-1} ,\quad |x| < 1$ where $\zeta(\cdot)$ denotes the Riemann-Zeta function . Thus, the first few terms of the Maclaurin series of $\Psi(1-z)$ is $-\gamma(z) - \zeta(2) z + \zeta(3) z^2 + z(\cdots)$ Putting it all together, the limit in question is $\lim_{x\to 0}\Gamma(x)(\gamma+\psi(1-x)) = \lim_{z\to 0} \left [ \frac 1z + \gamma + z( \cdots) \right ] \cdot \left [ - \zeta(2) z + \zeta(3) z^2 + z(\cdots) \right ] = -\zeta(2) = -\frac{\pi^2}6 \approx -1.64493.$