‎prove that the sequence ‎$‎\{F(n)\}‎$ ‎converges.‎ ‎‎‎‎Let ‎‎$‎g:‎\mathbb{R^+}‎‎‎\rightarrow‎‎\mathbb{R^+}‎$ ‎be a function such that $\log g(x)‎$ ‎is ‎concave, and‎ ‎$‎‎‎‎\displaystyle{\lim_{x\to\infty}}‎\frac{g(x+w)}{g(x)} = 1‎$ ‎‎‎‎‎‎‎‎‎for each ‎$‎w>0‎$ . ‎‎‎Then‎:‎
Fact 1: ‎‎$‎g(x)‎$ ‎is ‎increasing‎;‎
‎‎
Fact 2: ‎‎$\log g(x)‎$‎‎‎‎‎ has derivative ‎‎$‎‎\frac{g^\prime_{-}(x) + g^\prime_{+}(x) ‎}{2g(x)}‎$ ‎except, possibly, on a countable set, where ‎$‎g^\prime_{+}(x‏)‎$ ‎and ‎‎$‎g^\prime_{-}(x)‎$ ‎are ‎right ‎and ‎left ‎derivatives, ‎respectively; ‎
Fact  ‎3:‎‎ $‎‎\frac{g^‎\prime_{-}(x) + g^‎\prime_{+}(x) ‎}{2g(x)}‎$ is ‎decreasing ‎and ‎non-negative on ‎$‎‎\mathbb{R}‎^+‎$‎.‎
‎‎
My question ‎is:‎
‎‎Let ‎‎
‎‎\begin{align*}‎‎
‎F(n) = \sum_{i=1}^n ‎‎\frac{g^\prime_{-}(i) + g^\prime_{+}(i) ‎}{2g(i)} - \log g(n),
‎\end{align*}‎‎‎
‎prove that the sequence ‎$‎\{F(n)\}‎$ ‎converges.‎‎
‎Thanks‎ in advance.
 A: Alternative solution:
Fact 1: Let $f: \mathbb{R} \to \mathbb{R}$ be a convex function. 
Then, $f(x) - f(y) \ge c(x-y)$ for any real numbers $x, y\in \mathbb{R}$ and $c\in [f'_{-}(y), f'_{+}(y)]$.
See: https://en.wikipedia.org/wiki/Subderivative
https://www.sintef.no/globalassets/project/evitameeting/fho.pdf
A proof of Fact 1 is also given later.
From Fact 1, by letting $f = -\log g$ and $c = \frac{f'_{-}(y) + f'_{+}(y)}{2}$, we have,
for any $x, y\in (0, \infty)$, 
$$-\log g(x) + \log g(y) \ge -\frac{g'_{-}(y) + g'_{+}(y)}{2g(y)}(x-y). \tag{1}$$
By letting $x = i, y = i+1$; $x = i+1, y = i$ respectively, we have, for $i = 1, 2, \cdots$,
$$\frac{g'_{-}(i+1) + g'_{+}(i+1)}{2g(i+1)} \le \log g(i+1) - \log g(i) \le
\frac{g'_{-}(i) + g'_{+}(i)}{2g(i)}. \tag{2}$$
Let $n < m$ be two positive integers.
From (2), we have
$$\sum_{i=n}^{m-1} \frac{g'_{-}(i+1) + g'_{+}(i+1)}{2g(i+1)} \le \sum_{i=n}^{m-1} (\log g(i+1) - \log g(i)) \tag{3}$$
and
$$\sum_{i=n+1}^{m} (\log g(i+1) - \log g(i)) \le \sum_{i=n+1}^{m} \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)} \tag{4}$$
which results in
$$\log \frac{g(m+1)}{g(m)} - \log\frac{g(n+1)}{g(n)} \le \sum_{i=n+1}^{m} \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)} - \log\frac{g(m)}{g(n)} \le 0. \tag{5}$$
Thus, we have
$$\log \frac{g(m+1)}{g(m)} - \log\frac{g(n+1)}{g(n)} \le F(m) - F(n) \le 0.\tag{6}$$
Since $\lim_{k\to \infty} \log \frac{g(k+1)}{g(k)} = 0$, we know that
$\{F(n)\}$ is a Cauchy sequence. Thus, $\{F(n)\}$ has a limit denoted by $\gamma_g$.
Then, let us prove that
$$0 \le \gamma_g + \log g(1) \le \frac{g'_{-}(1) + g'_{+}(1)}{2g(1)}. \tag{7}$$
From (2), we have
$$\sum_{i=1}^{n-1} \frac{g'_{-}(i+1) + g'_{+}(i+1)}{2g(i+1)} \le \sum_{i=1}^{n-1} (\log g(i+1) - \log g(i))\tag{8}$$
and
$$\sum_{i=1}^{n} (\log g(i+1) - \log g(i)) \le \sum_{i=1}^n \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)}\tag{9}$$
which results in
$$\log\frac{g(n+1)}{g(n)} \le \sum_{i=1}^n \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)} - \log g(n) + \log g(1) \le 
\frac{g'_{-}(1) + g'_{+}(1)}{2g(1)}. \tag{10}$$
By using $\lim_{n\to \infty} \log\frac{g(n+1)}{g(n)} = 0$, the desired result follows.
$\phantom{2}$
Proof of Fact 1: We have, for any $r < t < y < u < v$,
$$\frac{f(r) - f(y)}{r- y} \le \frac{f(t)-f(y)}{t-y} \le \frac{f(u)-f(y)}{u-y} \le \frac{f(v)-f(y)}{v-y}.$$
Indeed, for example, $\frac{f(r) - f(y)}{r- y} \le \frac{f(t)-f(y)}{t-y}$ can be written
as $f(t) = f(\frac{t-r}{y-r}\cdot y + \frac{y-t}{y-r}\cdot r) \le \frac{t-r}{y-r}f(y) + \frac{y-t}{y-r}f(r)$ which is true by the definition of convex functions.
Thus, $f'_{+}(y) = \lim_{x \to y+} \frac{f(x)-f(y)}{x-y}$ and $f'_{-}(y) = \lim_{x \to y-} \frac{f(x)-f(y)}{x-y}$ both exist, 
and $f'_{-}(y) \le f'_{+}(y)$; moreover, if $x > y$, we have $f'_{+}(y) \le \frac{f(x)-f(y)}{x-y}$
which results in $f(x) - f(y) \ge f'_{+}(y)(x-y) \ge c(x-y)$ for any $c\in [f'_{-}(y), f'_{+}(y)]$,
and if $x < y$,
we have $\frac{f(x)-f(y)}{x-y} \le f'_{-}(y)$ which results in $f(x)-f(y) \ge f'_{-}(y)(x-y) \ge c(x-y)$
for any $c\in [f'_{-}(y), f'_{+}(y)]$. We are done.
A: The following can give ideas (and is too long for a comment).
You can bound the sum from below and above :
\begin{align*}
&&\int_1^{n+1} ‎‎\frac{g'_{-}(x) + g'_{+}(x) ‎}{2g(x)} dx&\leq\sum_{i=1}^n ‎‎\frac{g'_{-}(i) + g'_{+}(i) ‎}{2g(i)} \leq \int_0^n ‎‎\frac{g'_{-}(x) + g'_{+}(x) ‎}{2g(x)} dx\\
\Leftrightarrow &&\log g(n+1)-\log g(n)-\log g(1) &\leq F(n)\leq \log g(0)\\
\Leftrightarrow &&\log \frac{g(n+1)}{g(n)}-\log g(1) &\leq F(n)\leq -\log g(0)
\end{align*}
so when $n$ tends to infinity the left term tends to $0$. This do not guaranty convergence as is but maybe be of help.
