Question

What is the interval of convergence of the MacClaurin series of `f(x)=1 / (3-2x)`?

Answer 1

`1/(3-2x) = 1/3 sum_(n=0)^oo ((2x)/3)^n` for `x in [-3/2,3/2)`

Explanation:

To determine the MacLaurin's series for:

`f(x) = 1/(3-2x)`

we need to evaluate the derivatives of `f(x)` for all orders.

`d/dx 1/(3-2x) = d/dx (3-2x)^(-1) = (-2)(-1)(3-2x)^(-2) = 2/(3-2x)^2`

`d^2/dx^2 1/(3-2x) = d/dx 2(3-2x)^(-2) = (-2)(-2)2(3-2x)^(-3) = 8/(3-2x)^3`

and we can see that in general:

`d^n/dx^n 1/(3-2x) = (2^n n! )/(3-2x)^(n+1)`

so:

`[d^n/dx^n 1/(3-2x)]_(x=0) = (2^n n! )/(3^(n+1))`

So the MacLaurin series is:

`1/(3-2x) = sum_(n=0)^oo 1/3 (2/3)^n n! x^n/(n!) = 1/3 sum_(n=0)^oo ((2x)/3)^n`

This is a geometric series of ratio `r= (2x)/3` and so it is convergent for:

`-1 <= (2x)/3 < 1`

or:

`-3/2 <= x < 3/2`