Find the $Θ$ of $∑\log\log k$?

Question

I am struggling finding the $$\sum_{k=1}^{n}\log \log k$$

I know that $∑_{k=1}^n \log k = \log(n!) \implies Θ(n\log(n))$.

But unfortunately that does me help me to find the $Θ$ for $∑_{k=1}^n \log \log k$.

I can do not get a faculty to estimate the lower and upper bound . I think I have to use the rules of calculation of the log. So instead of having two sums, having one sum and one product, but that does really bring me closer to find an estimation of the upper or lower bound for the sum.

Edit: Thank you for the comments

$\log \log k$ with $k=1$ is undefined. Wolfram alpha says $ \log \log k $ at $k = 1 = -∞$. In the question it's asked from $k = 1$ to $n$

If it is undefined, there is no Θ I guess?

If it is -∞, I guess there is no Θ neither, since you can not really bound it?

And does anyone know the solution if the sum starts at $k=2$?

Answer 1

In general, if $f(n)$ is positive, monotonic increasing, and unbounded, and defined for $n \ge n_0$, if $f(n)$ increases slowly enough, then $\sum_{n_0 \le k \le n} f(k) =\Theta(nf(n)) $.

This is off the top of my head, but I think a sufficient slow growth condition would be $\lim\sup_{n \to \infty} \dfrac{f(an)}{f(n)} \le 1$ for some $a > 1$.

For $f(n) = \log(n)$ or $f(n) = \log\log(n)$ or more iteratedly (wasn't a word, is now) logs, $a = 2$ works.

Answer 2

$$n \log \log n \geq \sum_{k=2}^n \log \log k \geq \sum_{k=\frac{n}{2}}^n \log \log k \geq \frac{n}{2} \log \log \frac{n}{2}$$ The rightmost step is taking the minimal term and multiplying it by the number of terms. Note that

$$\log \log \frac{n}{2} = \log(\log n - \log 2) = \Theta(\log \log n)$$ And so we get the sum is $\Theta(n \log \log n)$. I'll let you fill in the formal details regarding the $\frac{n}{2}$, and proving the former bound.