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I was given the following thing to prove:

$$\lim_{n \to \infty} {d(n) \over n} = 0$$ where $d(n)$ is the number of divisors of n.

I'm so sure how to approach this question. One way I thought of is to use the UFT to turn the expression to:

$$\lim_{n \to \infty} {\prod (x_i + 1) \over \prod p_i^{x_i}}$$

And then to use L'Hôpital's rule for each $x_i$, so I get something like this:

$$\lim_{n \to \infty} {1 \over \ln (\sum p_i) \prod p_i^{x_i}}$$

That equals zero.

  1. Is this a good approach?
  2. Is there a different way to solve this?
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    $\begingroup$ For some $n$, the divisors can be at most: 1, 2, 3, 4 up to the smallest integer greater than $\sqrt(n)$, along with their correspondants n/1, n/2, etc.., i.e. there is at most $2\sqrt(n)$, and so the limit is bounded by $1/2/\sqrt(n)$ which converge to $0$ $\endgroup$ – Coolwater Feb 2 '16 at 18:46
  • $\begingroup$ following your argument, it seems possible to say that $2^k / \prod_{i=1}^k p_i \to 0$ and $(k+1) / p_i^k \to 0$ when $k \to \infty$ so together with $d(n) \le n$ it should prove that $\prod (x_i+1)/\prod p_i^{x_i} \to 0$ when $\prod p_i^{x_i} \to \infty$ $\endgroup$ – reuns Feb 2 '16 at 19:53
  • $\begingroup$ What is "UFT"?​ $\endgroup$ – Peter Mortensen Feb 2 '16 at 22:19
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    $\begingroup$ @Peter Mortensen : after some investigation, I think it stands for "unique factorization theorem"' $\endgroup$ – reuns Feb 2 '16 at 22:41
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It is actually true that $\lim_{n\to\infty} \frac{d(n)}{n^c}=0$ for any $c>0$, but if you only want it for $c=1$, the simplest proof would be the following. The divisors $\delta\mid n$ can be organized into pairs $(\delta,n/\delta)$, and in every pair the smallest divisor is $\min\{\delta,n/\delta\}\le\sqrt n$. It follows that the number of pairs is at most $\sqrt n$. Thus $d(n)\le 2\sqrt n$ and the assertion follows.

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    $\begingroup$ In case you are interested, here are two proofs of the fact you state at the beginning of the answer. $\endgroup$ – wythagoras Feb 2 '16 at 18:57
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    $\begingroup$ This would get down to $c = 1/2 + \epsilon$ as well. $\endgroup$ – Michael Lugo Feb 2 '16 at 19:22
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Hint: For every divisior $p$ greater than $\sqrt{n}$ of $n$, there is a divisor $q$ smaller than $\sqrt{n}$ such that $n=pq$. It follows that we can divide the divisiors in pairs where one of the elements is smaller than $\sqrt{n}$, and hence $d(n) \leq 2\sqrt{n}$. Can you use this to evaluate the limit?

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