Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This problem is taken from Vojtěch Jarník International Mathematical Competition 2010, Category I, Problem 1. — edit by KennyTM

On going through this post i happened to get the following 2 problems into my mind:

Let $f: \mathbb{N} \to \mathbb{N}$ be a bijection. Then does the series $$\sum\limits_{n=1}^{\infty} \frac{1}{nf(n)}$$ converge?

Next, consider the series $$\sum\limits_{n=1}^{\infty} \frac{1}{n+f(n)}$$ where $f: \mathbb{N} \to \mathbb{N}$ is a bijection. Clearly by taking $f(n)=n$ we see that the series is divergent. Does there exist a bijection such that the sum above is convergent?

share|cite|improve this question
Clearly, for the second question, if $f(n)$ satisfies $\lim_{n\to\infty}\frac{n}{f(n)}=0$ ($f(n)$ grows much faster than $n$), the second sum converges. – J. M. Sep 8 '10 at 9:42
J.M. It won't, since $f$ is a bijection, $n/f(n)\ge1$ infinitely often. – Robin Chapman Sep 8 '10 at 9:55
@Robin, mini-tip: when writing a comment to someone (even a commenter) you can prefix the name with an @ and s/he'll get a notice about it---like you about this one. – Mariano Suárez-Alvarez Sep 8 '10 at 10:07
@Robin: Hmm, on second thought, you're right. @Mariano: For me, sometimes it works, sometimes it doesn't. – J. M. Sep 8 '10 at 10:57
@Mariano: does one have to use the whole username, or @(user-first-name) suffices? e.g., are you getting a notice from this comment? – T.. Sep 8 '10 at 17:55
up vote 9 down vote accepted

Hints. For the first series, prove that $$\sum_{n=1}^N\frac1{nf(n)}\le\sum_{n=1}^N\frac1{n^2}.$$

For the second, what would happen if $f$ were an involution, and in each pair $(n,f(n))$ one term was much bigger than the other?

share|cite|improve this answer

A couple of proofs in italian.

share|cite|improve this answer
The problem comes from the 2010 Vojtěch Jarník International Mathematical Competition ( ). – Fosco Loregian Sep 8 '10 at 9:34

For question $2$, consider any $f(n)$ which is $2^n$ for $n$ which are not powers of $2.$ This lets us divide our sum into two parts, both of which converge.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.