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 $n\to +\infty$?