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Let $f$ be a real function defined on $[1, +\infty)$ and convex from a number on.

Is it true that the sequence $f_n:=f(n)$ is unbounded?

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    What about bounded above or below?( none-constant) – 2017-02-25
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    It is clear for constant, what about nonconstant functions? – 2017-02-25
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    What about $f(x)=\frac{1}{x}$? – 2017-02-25

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The answer is no. Indeed, choose a constant function $f(x)=1$ for all $x\in[1,+\infty]$. It is convex and bounded.

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    No, things aren't "different" in the strict case. The answer is still no. – 2017-02-25
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    @MathematicsStudent1122 yes indeed... – 2017-02-25