Does there exist a uniformly continuous function $f: (0,\infty) \to (0,\infty)$ such that $$\sum_{n \in \mathbb N} \frac{1}{f(n)}$$ converges?
Actually, I have tried some examples but they are against this statement, and obviously here $f(n)$ diverges to $+\infty$ for larger $n$, but after applying the Cauchy Criterion, I can't approach further.
No. There exist $\delta >0$ such that $|f(x) -f(y)| <1$ for whenever $|x-y|<\delta$. Dividing $[n, n+1]$ onto $n$ equal parts with $\frac 1 N <\delta$ we see that $|f(n+1)-f(n)| <N$ for all $n$ which implies that $f(n) < f(1)+nN$. Can yo see now why the series diverges?