Let $f:[1,+\infty)\rightarrow\mathbb{R}$ be a concave function such that $\displaystyle{\lim_{n\to\infty}}(f(n) - f(n+1)) = 0$ ( $n$ is positive integer). My question is :
What other condition is required that $\displaystyle{\lim_{x\to\infty}}f(x) = L$?
I know that if $\displaystyle{\lim_{x\to\infty}}f(x) = L$, then $\displaystyle{\lim_{n\to\infty}}f(n)= L$. Therefore $\displaystyle{\lim_{n\to\infty}}(f(n) - f(n+1)) = 0$.
And also, if $\displaystyle{\lim_{n\to\infty}}f(n) = L$ and $f$ is monotone, then $\displaystyle{\lim_{x\to\infty}}f(x) = L$. But $\displaystyle{\lim_{n\to\infty}}(f(n) - f(n+1)) = 0 \nRightarrow\displaystyle{\lim_{n\to\infty}}f(n) = L$.
Concavity is important here. That $f$ is concave means that $f$ is at or above any chord. That is, on any interval $[a,b]$, $$f(x) \ge \frac{f(a)(b - x) + (x-a)f(b)}{b - a}$$ In particular, $f\left(\frac{a+b}2\right) \ge \frac{f(a)+f(b)}2$. If we apply this to $a = n, b = n+2$, we get $$2f(n+1) \ge f(n) + f(n+2)\\f(n+1) - f(n+2) \ge f(n) - f(n+1)$$ By induction, if $n\ge m$, $$f(n) - f(n+1) \ge f(m) - f(m+1)$$
If for some $m, f(m) - f(m+1) = \epsilon > 0$, then for all $n > m, f(n)-f(n+1) \ge \epsilon$ as well, contradicting that $\lim_n f(n) - f(n+1) = 0$. That is, $f$ must be increasing (or at least constant) on the integers.
Now let $n > x_2 > x_1$ for some integer $n$ and real numbers $x_2, x_1$. We know that $f(n+1) \ge f(n)$. If $f(x_1) > f(n)$, then $f(n)$ would be below the line connecting $(x_1,f(x_1))$ to $(n+1,f(n+1))$, which cannot be. Therefore $f(x_1) \le f(n)$. If $f(x_1) > f(x_2)$, then $f(x_2)$ would be below the line connecting $(x_1,f(x_1)$ to $(n,f(n))$, which also cannot be. Therefore we can conclude that $f(x_2) \ge f(x_1)$. I.e., $f$ must be increasing.
And there you have it. Since $f$ is increasing, it has a limit if and only it is bounded above. Boundedness is the condition you are after.
Similarly, if you replace "concave" with "convex", $f$ will be decreasing, and have a limit if and only if it is bounded below.