Is there a $f : \mathbb{R}_+ \to \mathbb{R}_+$ such that :

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

  2. 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)

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    $\begingroup$ Nope, the question is to find (or prove there is not) such a $ f $ $\endgroup$ – M.LTA Dec 23 '15 at 23:09
  • $\begingroup$ @math635 The question was whether there exists an $f$ with the given property - saying $f(x)=x$ doesn't work doesn't say much about that. $\endgroup$ – David C. Ullrich Dec 23 '15 at 23:10
  • $\begingroup$ My first thought is maybe something like $f(x) = \log(x+1)$, because it works for $x_{n} = n$, but I don't know if it holds in general. $\endgroup$ – DMcMor Dec 23 '15 at 23:18

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 $f(x_n')\ge f(x_n)$.)

  • $\begingroup$ shouldn't the $x_i$'s be increasing? $\endgroup$ – math635 Dec 23 '15 at 23:56
  • $\begingroup$ I missed that. Doesn't matter... $\endgroup$ – David C. Ullrich Dec 23 '15 at 23:57
  • $\begingroup$ Yes I meant non decreasing but still not the main part of the question. You did the job ! Elegant solution ! $\endgroup$ – M.LTA Dec 24 '15 at 0:24

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