Monotonicity of the sequences of the lower and upper Darboux sums

Question

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.

Answer 1

Define measures $\mu^+_n=\frac 1n\sum_{i=1}^n \delta_{i/n}$ and $\mu^-_n=\frac 1n\sum_{i=0}^n \delta_{i/n}$. I claim that $\mu^+_n$ stochastically dominates $\mu^+_{n+1}$ to second order, and similarly, $\mu^-_n$ is stochastically dominated by $\mu^-_{n+1}$ to second order.

Checking this amounts to checking an inequality for the cumulative integrals of the cumulative distribution functions, as explained in the wiki article. The functions $S_n(x)$ and $s_n(x)$ are $\int f(t)\,d\mu^+_n(t)$ and $\int f(t)\,d\mu^-_n(t)$ respectively, where $f(t)$ is the convex decreasing function $f(t)=1/(x+t)$. Of course, $-f$ is then a concave increasing function and your desired inequality is a known consequence (see the wiki article again) of second order stochastic dominance.

Answer 2

This is a long comment to @AnthonyQuas's solution. It works out the details of the stochastic dominance as stated by Anthony.

Let $F_n(x) = \int_{-\infty}^x \,d\mu^+_n(t)$ be the cumulative distribution function of $\mu^+_n$, and $I_n(x) = \int_{-\infty}^x F_n(t)\,d\mu^+_n(t)$. $F_n$ is a non-decreasing step function, and $I_n$ is a continuous, convex, piece-wise linear function, explicitly, $I_n(x) = \frac{1}{n}\ \sum_{k=1}^n (x - \frac{k}{n})^+$.

We want to show $I_{n+1}(x) \ge I_n(x)$ for all $x$.

For $x \le \frac{1}{n+1}$ both functions are 0.

$I_n(1) = \frac{1}{2}(1 - \frac{1}{n})$, which increases with $n$, so $I_n(1) < I_{n+1}(1)$.

Over the interval $[1,+\infty)$, $I_n(x)$ is linear, with slope 1, so for $x > 1$, $I_n(x) = (x-1) + I_n(1) < (x-1)+I_{n+1}(1) = I_{n+1}(x)$.

Since $I_n$ is convex, and $I_{n+1}$ is piece-wise linear on the interval $[\frac{U}{n+1},\frac{U+1}{n+1}] \,\,(U=1,2,..,n)$, to show the inequality over the intervals, it suffices to show that it holds at the endpoints, i.e.: it suffices to show $I_{n+1}(\frac{U}{n+1}) \ge I_n(\frac{U}{n+1})$ for $U=1, 2, .., n+1$. The cases $U=1$ and $U=n+1$ have been considered above.

For $U=2,.., n$, a straightforward computation yields $I_{n+1}(\frac{U}{n+1})=\frac{1}{(n+1)^2} \frac{U(U-1)}{2}$.

We have, $I_n(\frac{U}{n+1}) = \frac{1}{n}\ \sum_{k=1}^n (\frac{U}{n+1} - \frac{k}{n})^+$. Note that for the range of $U$ under consideration, and for $k=1,..,n$,
$\frac{U}{n+1} - \frac{k}{n} > 0 \iff U-k > \frac{k}{n} \iff U-k \ge 1$
therefore, $I_n(\frac{U}{n+1}) = \frac{1}{n+1}\ \sum_{k=1}^{U-1} (\frac{U}{n+1} - \frac{k}{n})=$
$=\frac{n-1}{n^2(n+1)} \frac{U(U-1)}{2} = I_{n+1}(\frac{U}{n+1})(1-\frac{1}{n^2}) $, so the inequality holds at $\frac{U}{n+1}$.