Consider the following lower and upper Darboux sums $$ s_n(x)\ :=\ \sum_{k=1}^n\frac 1{n\cdot x+k} $$ and $$ S_n(x)\ :=\ \sum_{k=0}^{n-1}\frac 1{n\cdot x+k} $$

for every real $\ x>0\ $ and natural $\ n=1\ 2\ \ldots,\ $ so that

1. $\quad s_n\ <\ \log\frac{x+1}x\ <\ S_n $
2. $\quad S_n\ - s_n\ =\ \frac 1{n\cdot x\cdot(x+1)} $
3. $\quad s_{a}\ <\ s_{a\cdot b}\ \ <\,\ \ S_{a\cdot b}\ <\ S_{a} $

for every real $x>0$ and natural $\ n\ a\ b\ $ such that $\ b\ge2\ $ (these properties show that indeed there is a limit in the sense of Cantor--it's not even necessary to mention $\ \log\frac {x+1}x$).

Conjecture

$$ \forall_{n=1\ 2\ \ldots}\quad s_n<s_{n+1}\ <\,\ S_{n+1}<S_n $$

It'd be nice to have three proofs: (a) finite--a proof which doesn't mention any integers larger than n+2, nor any limit; (b) takes advantage of arbitrarily large natural numbers, and of nothing more; (3) takes advantage of function logarithm (in a useful way of course).

In general:

CONJECTURE'   Similar Darboux inequalities hold for arbitrary monotone and convex (or concave) real functions in one variable, with respect to partitions of a fixed interval of arguments into $n$ equal intervals.

For monotone continuous functions the conjecture would be wrong, there'd be easy counterexamples.