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

In this answer to a question about a series, a theorem was stated:

if $A= \{a_i \}$ is a set such that $\sum_{i = 1}^{\infty} \frac{1}{a_i}$ converges, then $d(A) = 0$, where $d(A)$ is the natural density of the set.

My background in number theory is basically zero and all my attempts to prove this have been utterly unseccessful; would anyone please help me, or at least provide me with a hint to prove it?

share|cite|improve this question
up vote 5 down vote accepted

If the natural density of $A = \{a_i\}$ exists, then we can show that it must be zero.

Let $\displaystyle S_{n} = \frac{|A \cup [1,n]|}{n}$

Now $\displaystyle \{\frac{n}{a_n}\}$ is a subsequence of $S_{n}$ and so if the limit is $\displaystyle 2\delta > 0 $ then we have that for all $\displaystyle n > N_0$, $\displaystyle \frac{n}{a_n} > \delta$ and so $\displaystyle \frac{1}{a_n} > \frac{\delta}{n}$ for all $\displaystyle n > N_0$ and so $\displaystyle \sum \frac{1}{a_n}$ diverges.

The main problem is actually showing that the limit exists.

It is easy to show that $\liminf$ is zero: If the limit was $\displaystyle 2\delta > 0$ then for all $n > N_{0}$, $S_{n} > \delta$ and an argument similar to above works.

Now suppose $\displaystyle \limsup S_n = 2\delta > 0$. Then there is a subsequence $\displaystyle S_{N_1}, S_{N_2}, ..., S_{N_k}, \dots $ which converges to $\displaystyle 2\delta$.

Now we can choose the subsequence so that $\displaystyle S_{N_i} > \delta$ and $\displaystyle N_{k+1} > \frac{2N_{k}}{\delta}$

Now the number of elements of $\displaystyle A$ in the interval $\displaystyle (N_{k}, N_{k+1}]$ is atleast $\displaystyle \delta N_{k+1} - N_k \ge \delta N_{k+1} - \frac{\delta N_{k+1}}{2} \ge \frac{\delta N_{k+1}}{2}$ and so the sum of reciprocals in that interval is atleast $\displaystyle \frac{\delta N_{k+1}}{2} \frac{1}{N_{k+1}} = \displaystyle \frac{\delta}{2}$

And so the sum of reciprocals must diverge.

Hence $\displaystyle \limsup S_n = 0 = \liminf S_n$ and thus $\displaystyle \lim S_n = 0$ and thus the natural density is zero.

share|cite|improve this answer

Given a set $A = (\cup a_i) \subset N$ with positive lim-sup density, it is possible to partition $N$ into intervals on which $A$ has density $\geq d$. If the intervals grow in size at a suitable rate (for this problem, exponential growth is enough), there will be a positive number $C>0$ such that on each interval, $\Sigma_{a \in A} 1/a \geq C$, and this makes the sum over all intervals infinite.

In more detail,

  1. There is a partition of the positive integers into intervals $[1,n_1], [n_1 + 1, n_2], [n_2 + 1,n_3], \dots$ such that the density of $A$ is at least $d > 0$ on each interval. This can be done for any $d < L$ where $L$ is the lim sup density of $A$.

  2. By joining together adjacent intervals, one can also require that the sequence $n_k$ grows fast enough that density in $[1,n_i]$ is not strongly affected by what happens in preceding intervals $[1,n_j]$ with $j < i$. In particular, one can keep or remove the subinterval $[1,n_{i-1}]$ with small effect on density questions in $[1,n_i]$. For this problem, fast enough growth will mean strictly exponential growth, i.e., $n_{i+1} / n_i > p $ for constant $p > 1$.

  3. The most efficient packing of a given number of integers into an interval $[m,n]$ (where efficiency means minimim sum of $1/a$ with $a$ ranging over the subset) is to place the whole subset as close as possible to $n$. If the density is at least $d$, the number of integers to be placed is at least $d(n-m+1)$, and then $\Sigma 1/a$ is at least $d(n-m)/n = d(1-m/n)$.

  4. For the exponentially growing intervals with $m/n < 1/p$ the sum in each interval of the partition of $N$ is bounded below by $C=d(1-1/p) > 0$, so the sum of all $1/a_i$ is infinite.

share|cite|improve this answer

EDIT nr 2: Please forget about this answer; as pointed out in the comments, it wasn't correct! (I've removed the text it here; see the edit history for the embarrasing details. Next time, I'll try to think properly before I post...)

share|cite|improve this answer
Hmm, I was a bit sloppy there. The argument is not quite right, althought I think the basic idea works. I'll see if I can fix it before someone else gives a better answer. ;) – Hans Lundmark Oct 3 '10 at 12:41
There, that's better (I hope). – Hans Lundmark Oct 3 '10 at 13:01
Thanks for re-elaborating. I'll wait a bit still to accept the answer to see if somebody else wants to give it a shot. :) – Andy Oct 3 '10 at 14:58
I am not sure this is right. 1) Existence of $d(A)$ has to be shown. 2) What if $d(A)/2 - |A\cup[1,N]|/(N+k) < 0$?. – Aryabhata Oct 3 '10 at 19:56
1) Good point. 2) I thought I had ruled this out somehow, but looking back at what I wrote, I don't know what I was thinking! – Hans Lundmark Oct 4 '10 at 3:24

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.