If $‎{\lim_{n\to\infty}}(f(n) - f(n+1)) = 0‎$, then $‎{\lim_{x\to\infty}}f(x) = L‎$. ‎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 :‎
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‎‎‎What other condition is required ‏‎that $‎‎‎\displaystyle{\lim_{x\to\infty}}f(x) = L‎$‎‎‎‎‎‎‎‎‎‎‎‎‎‎?‎
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 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
$‎‎‎\displaystyle{\lim_{n\to\infty}}(f(n) - f(n+1)) = 0‎ ‎‎‎\nRightarrow‎‎‎‎‎\displaystyle{\lim_{n\to\infty}}f(n) = L‎$.
 A: 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.
A: if a function has limit, it is bounded, and the condition of $sup(-f(x))=-inff(x)$ and $inf(-f(x))=-supf(x)$ correspond respectively to convex and concave function.
however, if a monotone function satisfy both these two conditions, it will behave like a wave and turbulence around its domain, in this case, this function do not have limit!
thank you!
