Does $\lim_{n\rightarrow\infty} \sum_{k = 0}^{n}{f(\frac{k}{n}) }$ exist? Why? $ f: [0;1] \rightarrow (0; +\infty) $,
$\lim_{n\rightarrow\infty} \sum_{k = 0}^{n}{f(\frac{k}{n}) }$ appears to be non-existent (as the gut feeling tells me), since it's basically the sum of images of all possible rationals between 0 and 1. Even if a sequence such as $\lim_{n\rightarrow\infty} \sum_{k = 0}^{n}{f(\frac{1}{2^k}) }$ converges, for example, we'd still have infinitely many series such as this one added up. So, what do you think? If it does exist, can you provide the example of such a function, if not, why?

You can certainly find examples where $S(n)=\sum\limits_{k = 0}^{n}{f(\frac{k}{n}) }$ is bounded above for all $n$:
for example if  $f\left(\dfrac ab\right)=\dfrac1{b^3}$ where $\frac kn$ is written as $\frac{a}{b}$ with $a,b$ coprime integers and $b\ge 1$,
then the sum is bounded above by its supremum  of $1+ \dfrac{\zeta(2)}{\zeta(3)}\approx 2.368433$ approached when $n$ is a large factorial, and bounded below by its infimum of $2$ approached  when $n$ is a large prime.
So this points a general counter example using Daniel Schepler's comments:  consider each odd prime $p$, and you can say that $$S(p!) -S(p) \ge f(\tfrac12)$$ since every term in $S(p)$ is included in $S(p!)$, though $f(\tfrac12)>0$ is not in $S(p)$ while it is in $S(p!)$.
So $S(n)$ does not have Cauchy convergence and so does not have a finite limit.
