I was given this code: 
I was asked to determine its complexity, now you could see that we are doing $\log_i(n)$ "actions" in each iteration of $i$. The problem is that the answer here is $\Theta (n)$ and I can't seem to understand/nor prove this, so bassically what I want to prove is
$$\Theta(\sum_{k=2}^{n} \frac{\ln(n)}{\ln(k)}) =\Theta(n)$$
or just
$$\Theta(\ln(n)\sum_{k=2}^{n} \frac{1}{\ln(k)}) =\Theta(n)$$
I tried proving $$\frac{\ln(n)\sum_{k=2}^{n} \frac{1}{\ln(k)}}{n} \to 1$$, or that the difference goes to zero, but I couldn't. Keep in mind I've only taken real analysis and haven't taken any course with advanced complexity or whatever, but then again I'd be happy to see solutions to this even if I haven't seen your methods in class because we are not required to prove these things in class, but I also can't see the intuition for this.
Any help is greatly appreciated.
Argument is adapted from this question and answer here
Pretty much everything hinges on the summation term $$ \sum_{k=2}^n\frac{1}{\log(k)} \in \Theta\left(\int_2^n\frac{1}{\log(x)}dx\right) $$ which is the function of the associated integral. Lets just call this integral $$ f(n)=\int_2^n\frac{1}{\log(x)}dx. $$
Let $g(x)$ be defined as $$ g(n) = \frac{n}{\log(n)} $$
We now want to show that $f(n)\in\Theta\left(g(n)\right)$ by taking the limit $$ \lim_{n\rightarrow\infty} \frac{f(n)}{g(n)} < \infty. $$
Taking the derivative of the top and bottom terms to use L'Hopital's rule (and applying fundemental theorem of calculus) we get that $$ \lim_{n\rightarrow\infty}\frac{\frac{1}{\log(n)}}{\frac{\log(n)-1}{\log(n)^2}} = \lim_{n\rightarrow\infty}\frac{\log(n)}{\log(n)-1}=1. $$
Therefore we know that by transitivity of $\Theta$ that $$ \sum_{k=2}^n\frac{1}{\log(k)} \in \Theta\left(\frac{n}{\log(n)}\right). $$ Multiplying this by the $\log(n)$ term that we left out, we find that the original sum derived from the function $$ \sum_{k=2}^n\log_k(n)\in\Theta(n) $$
Q.E.D.