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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!

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3  
Isn't it obvious that $1\le d(n)\le n$? –  Robin Chapman Oct 18 '10 at 10:32
    
@Robin: Yes. Thats correct. –  anonymous Oct 18 '10 at 10:33
3  
Excellent, then isn't it obvious what the radius of convergence is? –  Robin Chapman Oct 18 '10 at 10:53
    
@Robin: Yes, sir. But i am wondering how can we deduce the second fact! –  anonymous Oct 18 '10 at 10:57

2 Answers 2

up vote 1 down vote accepted

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

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1  
You don't need to use $n^{1/n}\to 1$ :-) –  Robin Chapman Oct 18 '10 at 16:49
1  
@Robin: Yes but $n^{1/n} \to 1$ was my reason for posting :-) –  Aryabhata Oct 18 '10 at 16:52

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.

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