uniformly continuous function $f$ such that $\sum 1/f(n)$ is convergent? 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 
 A: 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.
A: Assume that $\sum 1/f$ converges. Then for some $N_k \in \mathbb{N}$, 
$$n > N_k \implies |1/f(n)| < 1/k \implies |f(n)| > k.$$
$f$ is uniformly continuous. Therefore, there exists $m \in \mathbb{N}$ such that $|x-y| < 1/m \implies |f(x)-f(y)| < 1$. By the triangle inequality, this implies $|f(n+1)-f(n)|<m$ for all $n$.
Then if $a, a+1 > N_{\lceil m/2 \rceil} = N'$ are integers such that WLOG $f(a) > 0$ and $f(a+1) < 0$, then $f(a) > k$ and $f(a+1) < -k$. Then $|f(a+1)-f(a)|> 2\lceil m/2 \rceil > m$, contradiction. Therefore, $f(n)$ must not change sign for integer $n > N'$.
So, assume WLOG that $f(n)$ is eventually positive. Note this implies for $n > N'$ and $r \in \mathbb{N}$, that $f(n+r) > f(n)+ rm$.
Write
$$\sum_{n=1}^{\infty} \frac{1}{f(n)} = \sum_{n=1}^{N'} \frac{1}{f(n)} + \sum_{n= N'+1}^{\infty} \frac{1}{f(n)}$$
Realize that for the following series whose terms are strictly positive,
$$\sum_{n= N'+1}^{\infty} \frac{1}{f(n)} > \sum_{n = 0}^{\infty} \frac{1}{f(N'+1) + nm},$$
and the RHS obviously diverges. Contradiction.
