I'm aware that by properties of logarithm $$\sum_{k=1}^n \ln (k) = \ln (n!)$$
My question is if $$\sum_{k=1}^n \ln^2 (k) = \ln^2 (n!)?$$
Because when I am verifying the value where $n = 5$, I get different result... maybe Im missing something... is there a formula that defines $$\sum_{k=1}^n \ln^2 (k)\text{ ?}$$
Here's my computation: $$\sum_{k=1}^n \ln^2 (k) = \ln^2 (n!)$$ $$\sum_{k=1}^5 \ln^2 (k) = \ln^2 (5!)$$ $$\ln^2(1)+\ln^2(2)+\ln^2(3)+\ln^2(4)+\ln^2(5) = \ln^2 (120)$$ $$0+0.48045+1.20694+1.92181+2.59029 = 22.92007$$ $6.199494$ is not equal to $22.92007$.
$\sum_{k=1}^n\ln^2k\ne\ln^2n!$ because
$$\ln^2(xy)=(\ln x+\ln y)^2\ne\ln^2x+\ln^2y.$$