# Borel Summation regularization

I try to sum the following sum $$\sum _{k=0}^{\infty } (-1)^k k! \log (k+1)$$ using$$\sum _{k=0}^{\infty } (-1)^k k! \log (k+1)=\text{NIntegrate}\left[\frac{e^{-x} \text{PolyLog}^{(1,0)}(0,-x)}{x},\{x,0,\infty \},\text{WorkingPrecision}\to 50\right]$$ Mathematica do not gives answer Knowing that $$\sum _{k=0}^{\infty } (-1)^k x^k \log (k+1)=\frac{\text{PolyLog}^{(1,0)}(0,-x)}{x}$$ I try LevinU transform but do not work I think the value is $$\sum _{k=0}^{\infty } (-1)^k k! \log (k+1)=-0.17269850566423522146907597201$$ could you verify with other sum methods