How to prove that $\sum\limits_{n=1}^{\infty}f(\frac{1}{n})$ convergence? Let $f:\mathbb{R}\to\mathbb{R}$ be convex and $\limsup\limits_{x\to0^+}\frac{f(\frac{x}{2})}{f(x)}<\frac{1}{2}$.Prove that $\sum\limits_{n=1}^{\infty}f(\frac{1}{n})$ convergence.This is our final exam last term.All I know is that when $n$ is large enough, $f(\frac{1}{2n})\leqslant\frac{1}{2}f(\frac{1}{n})$ holds.But I don't know how to use the convexity of $f(x)$.
 A: Let $q$ be a number with $\limsup_{x\to 0^+} \frac{f(x/2)}{f(x)}<q<\frac12$. For the $\limsup$ to even make sense, there must exist some $\epsilon>0$ such that for all $0<x<\epsilon$, we have
$ f(x)\ne 0$ and $\frac{f(x/2)}{f(x)}<q$.
Convex implies continuous. Suppose $f(0)\ne 0$. Then $\lim_{x\to 0}\frac{f(x/2)}{f(x)}=1$, contradicting $q<\frac12$. We conclude that $f(0)=0$.By convexity,
$$\tag1 f(\tfrac x2)\le\frac{f(x)+f(0)}{2}=\frac12f(x).$$
Note that $f$ has constant sign on $\left]0,\epsilon\right[$.
If $f(x)<0$ for $x$ in that interval, $(1)$ implies $q>\frac{f(x/2)}{f(x)}\ge \frac12$, contradiction. We conclude that $f$ is positive on $\left]0,\epsilon\right[$; moreover, $f$ is strictly increasing on that interval.
But if $f$ is positive, we find in fact that
$$ f(x/2)<qf(x)$$ for all $x\in\left]0,\epsilon\right[$.
Let $a_n=f(\tfrac1n)$. From the above, (possibly ignoiring finitely many initial exceptions) the sequence $(a_n)_n$ is positive and strictly decreasing and
$$ \sum_{n=2N}^{4N-1}a_n= \sum_{n=2}^{2N-1}(a_{2n}+a_{2n+1})\le  \sum_{n=N}^{2N-1}(2a_{2n})<2q\sum_{n=N}^{2N-1}(a_{n}).$$
Conclude.
