Suppose $f:(0,\infty)\rightarrow(0,\infty)$ is uniformly continuous. Then $\sum\frac{1}{f(n)}$ diverges? Suppose $f:(0,\infty)\rightarrow(0,\infty)$ is uniformly continuous. Then $\sum\frac{1}{f(n)}$ converges?
I have an example to $f(x)=x$. Which solves my question. But Can we say
Suppose $f:(0,\infty)\rightarrow(0,\infty)$ is uniformly continuous. Then $\sum\frac{1}{f(n)}$ diverges? ..............(1)
I think we can. 
It will be solved if we can prove that, $f:(0,\infty)\rightarrow(0,\infty)$ is uniformly continuous if $f(x)\leq cx+d$, where $c$ is a positive constant.......(2)
I don't the validity of the last statement. But I feel it is true. But I don't know. 
So my questions: Can I prove (2), if not, please provide a counterexample and how do we prove (1)? 
 A: In general, if $f$ is uniformly continuous, then $f(x)= O(x)$. This follows directly from definitions: if $\delta$ and $\epsilon$ are be positive reals so that $\lvert f(x) - f(y)\rvert \le \epsilon$ whenever $\lvert x - y \lvert \le \delta$, then for all $x > 0$ we have
$$\lvert f(x) \rvert \le \lvert f(x) - f(0) \rvert + \lvert f(0) \rvert \le \\ 
\lvert f(\delta) - f(0) \rvert + \lvert f(2 \delta) - f(\delta) \rvert + \ldots + \lvert f(x) - f(x - \left \lfloor \frac x \delta \right \rfloor \delta) \rvert + \lvert f(0) \rvert \\ \le \left \lceil \frac x \delta \right \rceil \epsilon + \lvert f(0) \rvert = O(x)$$
Now, this proves that $\sum \frac 1 {f(n)}$ cannot converge absolutely. Actually, it is also easy to see that, if $\lvert f(n)\rvert$ converges to infinity along natural $n$ (as must happen if our series is to converge), then $f$ must be constantly negative or positive after a certain value. Thus, the series would have to converge absolutely if it were to converge at all. Your suspicion is right; $\sum \frac 1{f(n)}$ always diverges.
