Prove or disprove: If $f(x)$ is continuous in $(0,1]$ and $f(x)\to\infty$ as $x\to 0^+$, then $\lim_{n\to\infty}\sum_{k=1}^n f(k/n)$ does not exist. I'm trying to prove or disprove the following conjecture:

If $f(x)$ is continuous in $(0,1]$ and $f(x)\to\infty$ as $x\to 0^+$ then $L=\lim\limits_{n\to\infty}\sum\limits_{k=1}^n f\left(\frac{k}{n}\right)$ does not exist.

My attempt
I tried proof by contradiction. Asssume $L$ exists.
$L=\lim\limits_{n\to\infty}n(\frac{1}{n})\sum\limits_{k=1}^n f\left(\frac{k}
{n}\right)=\left(\lim\limits_{n\to\infty}n\right)\int_{0}^1 f(x)dx$
(EDIT: As mentioned by @FShrike in the comments, the previous step is not valid.)
$\therefore \int_{0}^1 f(x)dx=0$
There are functions $f(x)$, continuous in $(0,1]$, such that $f(x)\to\infty$ as $x\to 0^+$ and $\int_{0}^1 f(x)dx=0$  . For example, $f(x)=-\ln{x}-1$, in which case $L$ does not exist, by Stirling's approximation.
But I do not see why all such functions $f(x)$ would imply that $L$ does not exist.
Context:
I am interested in geometrical infinite products (example1, example2). The conjecture in this question, via the substitution $f(x)=-\ln{g(x)}$, is equivalent to: If $g(x)$ is continuous in $(0,1]$ and $\lim\limits_{x\to 0^+}g(x)=0$ then $\lim\limits_{n\to\infty}\prod\limits_{k=1}^ng\left(\frac{k}{n}\right)$ either equals $0$ or does not exist, which stands in interesting contrast with the fact that infinite products of lengths or areas, that tend to $0$, can equal a positive number.
EDIT2:
I'm not sure if this is helpful, but I have noticed that $L_2=\lim\limits_{n\to\infty}\sum\limits_{k=1}^n f\left(\frac{k-1/2}{n}\right)$ can exist.
For example, $\lim\limits_{n\to\infty}\sum\limits_{k=1}^n \left(-\ln{\left(\frac{k-1/2}{n}\right)}-1\right)=-\frac{\ln{2}}{2}$. (Another question of mine yielded methods for dealing with the sum $\sum\limits_{k=1}^n \ln{(k-\frac12)}$.)
I do not understand why replacing $k$ with $k-\frac12$ seems to make the limit existable (if that's a word).

The statement in the title of this question can be disproved. I posted an equivalent question on Math Overflow, and it has been answered.
Note: The $f(x)$ in this question, and the $f(x)$ in the Overflow question, are different. Here is how they are related:
$$[f(x)\text{ in this question}]=-\ln{[f(x)\text{ in the Math Overflow question}]}$$
