TITLE: uniformly continuous function $f$ such that $\sum 1/f(n)$ is convergent?
QUESTION [8 upvotes]: Does there exist a uniformly continuous function $f:[1,\infty)\to \mathbb R$ such that $\sum_{n=1}^\infty 1/f(n)$ is convergent ?  
I know that $\exists M>0$ such that $|f(x)|< Mx, \forall x\in [1,\infty)$, so $|1/f(n)|>1/(Mn) ,\forall n \ge 1$, thus $\sum_{n=1}^\infty |1/f(n)|$ is divergent. But I don't know what happens with $\sum_{n=1}^\infty 1/f(n)$.
Please help

REPLY [8 votes]: Let us assume that such a function $f$ exists. From the uniform continuity we get a constant $M > 0$ such that
$$ \tag 1
 |f(n+1) - f(n)| < M
$$
for all $n \in \Bbb N$. In particular $|f(n)| < |f(1)| + nM$, so that
$\sum_{n=1}^\infty \frac{1}{|f(n)|}$ diverges (as you already observed).
If $\sum_{n=1}^\infty \frac{1}{f(n)}$ is convergent then
necessarily $|f(n)| \to \infty$ so that 
$$ \tag 2
 |f(n)| > M
$$
for $n \ge n_0$.
Combining these inequalities it follows that for $n \ge n_0$, all $f(n)$ have the same sign, so that $\sum_{n=1}^\infty \frac{1}{f(n)}$
is absolutely convergent, in contradiction to the above observation.
Therefore no such function $f$ exists.