# Rational function from series?

• Jun 3rd 2010, 03:14 PM
igopogo
Rational function from series?
Given the series:

$\displaystyle f(z)=\sum_{n=2}^{\infty} \frac{n(n-1)}{3^n}z^{n-2}$

a) find the radius of convergence (got that, it's z>1/3)

b) Prove that f is a rational function inside its circle of convergence

c) find the rational number f(2)

Well, I mess around with this for a while and get:

$\displaystyle f(z)=\sum_{n=2}^{\infty} \frac{n!}{3^2}(\frac{z}{3})^{n-2}\frac{1}{(n-2)!}$

Which is interesting, and sort of reminds me of $\displaystyle \frac{1}{e}$ and $\displaystyle e^z$, but a) I know multiplication of series doesn't work like that and b) that wouldn't be rational anyway. Any hints?
• Jun 3rd 2010, 03:26 PM
Drexel28
Quote:

Originally Posted by igopogo
Given the series:

$\displaystyle f(z)=\sum_{n=2}^{\infty} \frac{n(n-1)}{3^n}z^{n-2}$

a) find the radius of convergence (got that, it's z>1/3)

b) Prove that f is a rational function inside its circle of convergence

c) find the rational number f(2)

Well, I mess around with this for a while and get:

$\displaystyle f(z)=\sum_{n=2}^{\infty} \frac{n!}{3^2}(\frac{z}{3})^{n-2}\frac{1}{(n-2)!}$

Which is interesting, and sort of reminds me of $\displaystyle \frac{1}{e}$ and $\displaystyle e^z$, but a) I know multiplication of series doesn't work like that and b) that wouldn't be rational anyway. Any hints?

Two derivatives of a well known function.
• Jun 3rd 2010, 04:07 PM
igopogo
Nope, feeling dense here. Can you push me toward a *method* of looking at a problem like this? This is study, not homework
• Jun 3rd 2010, 06:54 PM
igopogo
nvm...solved. Thanks for the direction!