uniformly continuous function $f$ such that $\sum 1/f(n)$ is convergent?

Question

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

Answer 1

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.

Answer 2

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.