Show $\lim_{n\to\infty} \sum_{j=1}^n f(j) - F(n) - ‎\frac{1}{2}‎f(n) - c = 0$ where $f:[1,\infty]\to\Bbb{R}$ concave increasing and $F$ primitive Let the function ‎$‎f:[1, +‎\infty‎)‎‎\rightarrow‎‎\mathbb{R}‎‎‎$ ‎be ‎both ‎increasing ‎and ‎concave.  ‎Let ‎‎$‎F‎:‎[1, +‎\infty‎)‎‎\rightarrow‎‎\mathbb{R}‎$ ‎be a‎ ‎primitve ‎function ‎of ‎‎$‎f‎$‎. Then there exists ‎$‎c‎>0‎$ ‎such ‎that ‎‎
‎\begin{align*}‎‎
\displaystyle{\lim_{n\to\infty}}(\sum_{j=1}^n f(j) - F(n) - ‎\frac{1}{2}‎f(n) - c) = 0.
‎\end{align*}
Please guide me? thanks a lot.

Note that this question asks for the higher order terms of the Euler-Maclaurin Sum Formula in the case of increasing concave functions.
 A: Define
$$
a_n=\sum_{k=1}^nf(k)-F(n)-\frac12f(n)
$$
Because $f$ is concave,
$$
\begin{align}
a_n-a_{n-1}
&=\overbrace{\color{#C00}{\frac{f(n)+f(n-1)}2}}^\text{area under the secant}-\overbrace{\vphantom{\frac f2}\color{#00F}{(F(n)-F(n-1))}}^\text{area under the curve}\\
&\le0
\end{align}
$$
Thus, $a_n$ is decreasing.

Define
$$
b_n=\sum_{k=1}^nf(k)-F(n)-\frac12f(n+1)
$$
Because $f$ is concave,
$$
\begin{align}
b_n-b_{n-1}
&=\overbrace{\color{#090}{\frac{3f(n)-f(n+1)}2}}^{\substack{\text{area under the secant}\\\text{from the next interval}\\\text{extended over this interval}}}-\overbrace{\vphantom{\frac f2}\color{#00F}{(F(n)-F(n-1))}}^\text{area under the curve}\\
&\ge0
\end{align}
$$
Thus, $b_n$ is increasing.
Because $f$ is increasing, $a_n\ge b_n\ge b_0=-F(0)-\frac12f(1)$. Thus, $a_n$ is decreasing and bounded below, therefore, convergent.
Let
$$
c=\lim_{n\to\infty}a_n
$$
then
$$
\lim_{n\to\infty}\left(\sum_{k=1}^nf(k)-F(n)-\frac12f(n)-c\right)=0
$$
Whether $c$ is positive or not, depends on which primitve of $f$ we take for $F$.
A: Let $a_n$ be defined by $$
a_n=\sum_{j=1}^n f(j) -\frac12 f(n)-F(n).
$$ Then, we have
$$
a_{n}-a_{n+1} = -\frac{f(n+1)+f(n)}{2}+\int_n^{n+1}f(t)dt.
$$ Since for all $u\in [0,1]$
$$
f\left(un+(1-u)(n+1)\right)\ge uf(n)+(1-u)f(n+1)
$$ by concavity of $f$, we have
$$\begin{align*}
\int_n^{n+1}f(t)dt&=\int_0^1 f\left(un+(1-u)(n+1)\right)\mathrm d u\\
&\ge \int_0^1 uf(n)+(1-u)f(n+1)\mathrm d u\\
&=\frac{f(n)+f(n+1)}{2}.
\end{align*}$$ This implies $a_{n}-a_{n+1}\ge 0$ for every $n\ge 1$. Thus $a_n$ is decreasing.
On the other hand, concavity of $f$ implies that for all $u\in [n,n+1]$,
$$
f(n+1) \ge \frac{1}{n+2-u} f(u)+\frac{n+1-u}{n+2-u}f(n+2),
$$ or equivalently,
$$
f(u) \le (n+2-u)f(n+1)-(n+1-u)f(n+2).
$$ Integrating over $[n,n+1]$ yields:
$$
\int_n^{n+1} f(u)\mathrm du\le \frac32f(n+1)-\frac12 f(n+1),
$$ which implies
$$
a_n-a_{n+1}\le \frac{2f(n+1)-f(n)-f(n+2)}{2}.
$$ Summing $a_n-a_{n+1}$, we find that
$$\begin{align*}
0\le a_1-a_{N}&\le \sum_{n=1}^{N-1} \frac{2f(n+1)-f(n)-f(n+2)}{2}\\&=\frac{f(N)-f(N+1)+f(2)-f(1)}{2}\\
&\le \frac{f(2)-f(1)}{2}.
\end{align*}$$
By monotone convergence theorem, bounded decreasing sequence $(a_n)$ has a finite limit
$$
c=\lim_{n\to\infty} a_n.
$$
