# Convergence of $\sum\limits_{n=1}^{\infty} \frac{1}{nf(n)}$

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

On going through this post Does there exist a bijective $f:\mathbb{N} \to \mathbb{N}$ such that $\sum f(n)/n^2$ converges? 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?

• 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. isn't a mathematician 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-Álvarez 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. isn't a mathematician Sep 8 '10 at 10:57

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?
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.