Approximation rate for Greedy Set Cover algorithm (RESOLVED)
I have figured out the answer.

The trick is not really coming from $\frac{k}{t} &gt; ln(n)$ to $\frac{k}{t} \leq ln(n) + 1$ directly. Lets take 1 step back.


Lets say there is no item uncovered at k-th step. So, $|U_{k}| = 0$ and $n \times exp(-\frac{k}{t}) &lt; 1$ (otherwise, there could be a remaining uncovered item in $U_{k}$, and this would be a contradiction of $|U_{k}| = 0)$;


The step above can be rewritten as: 


Lets say there is no item uncovered at k-th step. So, $|U_{k}| = 0$ and we are done. We are interested in how much steps we can take before that happens. So, find the upper bound of k such as $|U_{k}| \geq 1$;
$|U_{k}| \geq 1 \Leftrightarrow n \times exp(-\frac{k}{k}) \geq 1$. This results in $k \leq \frac{ln(n)}{t}$. So, $k = \frac{ln(n)}{t}$ is our upper bound for iterations. If $k = \frac{ln(n)}{t}$, $|U_{k}| = 1$. It's almost done. There is only 1 last item to cover and, consequently, 1 last round to run. 
Therefore, $k \leq \frac{ln(n)}{t} + 1$ in order to get $|U_{k}| = 0$. It's easy to see that $\frac{k}{t} \leq ln(n) + \frac{1}{t}$. 


One can still get a cleaner solution and say $\frac{k}{t} \leq ln(n) + 1$ since $t \geq 1$.

k is the number of steps taking to get G_OPT. At each step, G_OPT increases by 1. Then, $|G\_OPT| = k$. We defined $t = |OPT|$. We end up with 

Approx. rate $= \frac{|G\_OPT|}{|OPT|} \leq ln(n) + 1$