Convergence of $\int_0^n f(x)\, dx - [\frac{1}{2} f(0) + f(1) +\dots + f(n-1) + \frac{1}{2} f(n)]$ If $f(x)$ is increasing, strictly concave, and twice continuously differentiable for $x > 0$, does the difference
$$
d_n = \int_0^n f(x)\, dx - [\tfrac{1}{2} f(0) + f(1) + f(2) +\dots + f(n-1) + \tfrac{1}{2} f(n)]
$$
converge to a finite positive limit when $n\to +\infty$?
The error form provided by the trapezoidal rule is not helpful since it just asserts that $d_n = -n f''(\xi_n)/12\geqslant 0$, for some $0 < \xi_n < n$, or that
$$
d_n\leqslant -\tfrac{1}{12}[f''(\xi_1) + f''(\xi_2) +\dots + f''(\xi_n)]
$$
with  $k-1 < \xi_k < k$. Since $f(x)$ is increasing and concave, then $f''(x)$ is negative, bounded, and increasing for $x > 0$, but then again, how to assure the converge of $\sum_{k=1}^n f''(\xi_n)$ when $n\to +\infty$?
 A: Consider
$$
f(x)=-\frac{1}{(x+1)(x+2)}.
$$
Clearly $f$ is increasing and concave in $[0,\infty)$.
Then
$$
\frac{1}{2}f(0)+f(1)+\cdots+f(n-1)+\frac{1}{2}f(n)=\cdots=-\frac{3}{4}+\frac{1}{n+1}-\frac{1}{2(n+1)(n+2)}\\=-\frac{3}{4}+\frac{1}{n}+{\mathcal O}(n^{-2})
$$
while
$$
\int_0^n f(x)\,dx=\int_0^n \frac{dx}{x+2}-\int_0^n\frac{dx}{x+1}=\log(n+2)-\log 2-\log(n+1)\\=-\log 2-\log\Big(1+\frac{1}{n+1}\Big)\\=-\log 2-\frac{1}{n}+{\mathcal O}(n^{-2})
$$
Thus
$$
d_n\to -\frac{1}{2}+\log 2>0.
$$
A: As you found out, the answer has to lie in the conflict field of "increasing" and "concave". Let's try a solution without referring to derivatives.
Concave: implies that for $x\in[a,b]$ one has that the function graph is above the secant,
$$
f(x)\ge \frac{(x-a)f(b) + (b-x)f(a)}{b-a}
\implies \int_n^{n+1} f(x)\, dx\ge \frac12 (f(n) + f(n+1)).
$$
Switching the points around, that is, exchanging $x$ and $b$ and again isolating $f(x)$ gives for $a < b < x$
$$
(x-a)f(b)\ge (b-a)f(x) + (x-b)f(a)\\
f(x)\le \frac{(x-a)f(b) + (b-x)f(a)}{b-a},
$$
which implies
\begin{align}
\int_{n+1}^{n+2} f(x)\, dx
&\le \int_{n+1}^{n+2}[(x-n)f(n+1) + (n+1-x)f(n)]\, dx\\
&=\frac12(3f(n+1)-f(n)),
\end{align}
so that
$$
0\le \int_n^{n+1} f(x)\, dx - \frac1 2(f(n)+f(n+1))\le \frac12 [2f(n)-f(n+1)-f(n-1)].
$$
Summing up from $m$ till $n$ we get
\begin{align}
0 &\le d_n-d_m\\
&= \int_m^n f(x)\, dx -\Bigl(\frac12 f(m)+ f(m+1) +\dots + f(n-1) +\frac12 f(n)\Bigr)\\
&\le \frac12 (f(m)-f(m-1)) -\frac12 (f(n) - f(n-1)). \tag{*}
\end{align}

From concavity we saw that $2f(n) - f(n+1) - f(n-1)\ge 0$. It immediately follows that the sequence $v_n = f(n+1) - f(n)$ is monotonically falling. Because $f$ is increasing, this sequence stays non-negative, so it has a non-negative limit $v$. This means that the sequence $v_n$ is a Cauchy sequence, and (*) then implies that also the sequence of the $d_n$ is Cauchy. Thus it has a limit, which can not be negative.
