1
$\begingroup$

I came across this question.

Let $d(n)$ denote the number of divisors of $n$. Let $$\nu(z) = \sum\limits_{n=1}^{\infty} d(n) z^{n}$$ Whats the radius of convergence of this power series. We also have to show that $$\nu(z) = \sum\limits_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}$$

Regarding the divisor function, we have Dirichlet's formula in hand. But will that help!

  • 3
    Isn't it obvious that $1\le d(n)\le n$?2010-10-18
  • 0
    @Robin: Yes. Thats correct.2010-10-18
  • 3
    Excellent, then isn't it obvious what the radius of convergence is?2010-10-18
  • 0
    @Robin: Yes, sir. But i am wondering how can we deduce the second fact!2010-10-18

2 Answers 2

1

Elaborating upon Robin's hint about the radius of convergence, we have that

$\displaystyle 1 \le d(n)^{1/n} \le n^{1/n}$

Since $\displaystyle n^{1/n} \to 1$, the radius of convergence of the power series is $\displaystyle 1$.

The reason I posted this is because I wanted to mention this quick proof of $\displaystyle n^{1/n} \to 1$

Let $\displaystyle n > 3$ then we have that

$\displaystyle \frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{n} \ge \sqrt[n]{n}$

using $AM \ge GM$ on $\displaystyle n-2 \ 1$s and two $\sqrt{n}$s.

Thus $\displaystyle \frac{n-2 + 2\sqrt{n}}{n} \ge n^{1/n} \ge 1$

i.e

$\displaystyle 1- \frac{2}{n} + \frac{2}{\sqrt{n}} \ge n^{1/n} \ge 1$

and so $\displaystyle n^{1/n} \to 1$.

  • 1
    You don't need to use $n^{1/n}\to 1$ :-)2010-10-18
  • 1
    @Robin: Yes but $n^{1/n} \to 1$ was my reason for posting :-)2010-10-18
2

Write your series as $$\sum_{n=1}^\infty \sum_{m=1}^\infty z^{mn},$$

and consider $mn=k,$ say. Does this help you?

If you get stuck look here.