Functions such that $\sum \frac{1}{x_n}$ diverges $\Longrightarrow \sum \frac{1}{x_nf(x_n)}$ diverge Is there a $f : \mathbb{R}_+ \to \mathbb{R}_+$ such that :


*

*$f$ is an increasing bijective map of $\mathbb{R}_+$ into itself.

*For all $\displaystyle\sum_n
\frac{1}{x_n}$ where $(x_n)$ is increasing and potitive :
$$\sum \frac{1}{x_n} \; \text{diverge}\; \Longrightarrow \sum \frac{1}{x_nf(x_n)} \; \text{diverge}$$
(From a French oral examination)
 A: Edit: What I wrote is not quite right. Or maybe it's right. See comment at bottom.
There is no such $f$. 
If $f$ is an increasing bijection there exists $y_j>j$ such that $f(y_j)>j$. Let $N_j$ be the smallest positive integer with $$N_j\frac1{y_j}>\frac1j.$$Since $1/y_j<1/j$ it follows that $$N_j\frac1{y_j}\le \frac2j.$$
Let $(x_n)$ be the sequence consisting of $y_1$ repeated $N_1$ times, followed by $y_2$ repeated $N_2$ times, etc. Then
$$\sum_n\frac1{x_n}=\sum_j N_j\frac1{y_j}>\sum_j\frac1j=\infty,$$while $$
\sum_n\frac1{x_nf(x_n)}=\sum_jN_j\frac1{y_j\,f(y_j)}\le2\sum_j\frac1{j^2}<\infty.$$
Comment: I missed the condition that the $x_n$ are supposed to be increasing. We can certainly make the $y_j$ increasing, in which case the $x_n$ are non-decreasing, which is what "increasing" often means. If we want the $x_n$ to be strictly increasing, start with $y_j$ strictly increasing, define $x_n$ as above, and then modify $x_n$ a tiny bit to make the sequence strictly increasing. If the modification is small enough this will not change the convergence or divergence of the two series. (For example, given $x_n$ as above we can certainly find a strictly increasing sequence $(x_n')$ such that $x_n\le x_n'\le 2x_n$; note that $f(x_n')\ge f(x_n)$.)
