Regularizing the $\log\log n$ series

Question

The divergent series $$\sum_{n=1}^\infty\log n$$ can be regularized using the derivative of the Riemann zeta function at $s=0$:

$$-\frac{\mathrm{d}}{\mathrm{d}s}\zeta(s)=-\frac{\mathrm{d}}{\mathrm{d}s}\sum_{n=1}^\infty n^{-s}=-\sum_{n=1}^\infty\frac{\mathrm{d}}{\mathrm{d}s}n^{-s}=\sum_{n=1}^\infty n^{-s}\log n$$

This yields $\frac{1}{2}\log(2\pi)$ as the answer. Is it possible to regularize the series $$\sum_{n=2}^\infty\log\log n$$ in a similar manner?

Answer 1

How about this: $$ f(s)=\sum_{n=2}^{\infty}e^{-s (\log n + \log\log n)}=\sum_{n=2}^{\infty}(n\log n)^{-s}, $$ so $$ f'(s)=-\sum_{n=2}^{\infty}(\log n + \log\log n)e^{-s(\log n + \log \log n)}, $$ and $$ \sum_{n=2}^{\infty}\log \log n = -f'(0) + \zeta'(0) = -f'(0)-\frac{1}{2}\log(2\pi). $$ Certainly $f(s)$ is analytic for $\Re(s)> 1$, so the question is just whether it can be analytically continued to $s=0$.

Answer 2

i make the admission up front that I am not an analytic number theory expert but what about approaching this by noting that the sum of divisors function $\sigma(n)=\sum_{d|n}d$ was shown by Gronwall in $1913$ to be written as $$\limsup \frac{\sigma(n)}{n \log \log n} = e^{\gamma}$$ Where $\gamma$ is Euler's constant.