# number of divisors of a composite number

Consider the power series $\sum_{n\ge1} a_n z^n$ where $a_n =$ number of divisors of $n^{50}$. then the radius of convergence of $\sum_{n\ge1} a_n z^n$ is

(1) 1

(2) 50

(3) $\frac 1 {50}$

(4) 0

-
So, what do you know about the number of divisors of $n^{50}$? –  Gerry Myerson Nov 16 '12 at 4:26
@Gerry, I see your point. –  Will Jagy Nov 16 '12 at 4:47
@Gerry if n is prime then the number of divisors of n^50 is 51, but when n is composite..? –  amritha Nov 16 '12 at 5:39
You don't need an exact answer, just a bound good enough to tell you when the series converges. –  Gerry Myerson Nov 16 '12 at 11:05

## 1 Answer

Hint: what is the radius of convergence of $\sum_{n\ge1}z^n$? of $\sum_{n\ge1}n^{50}z^n$?

-
@ Gerry 1 is the radius of convergence of ∑n≥1zn –  amritha Dec 15 '12 at 20:07
Yes, the radius of convergence of $\sum_{n\ge1}z^n$ is $1$. Can you work out the radius of convergence of the other sum I mention? maybe using the ratio test? –  Gerry Myerson Dec 15 '12 at 23:41