# Thread: Maclaurin Series, could you verify this?

1. ## Maclaurin Series, could you verify this?

Hi to you,

Q: Find the Maclaurin series of $f(x) = xe^x$.
Also find the associated radius of convergence.

A: So far I have found this for the first part:
$f^n(x) = e^x(n+x)$.

so it follows that:
$f^n(0) = n$.

and so:
$xe^x = \sum\frac{x^n}{(n-1)!}$ n going from 0 to infinity.

That is:
$xe^x = x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+....$

I am right for the first part?

Second part:

I did the ratio test and I got that the limit is smaller then 1.
Therefore the sum is absolutely convergent...???

Stev381

2. Originally Posted by Stev381
Hi to you,

Q: Find the Maclaurin series of $f(x) = xe^x$.
Also find the associated radius of convergence.

A: So far I have found this for the first part:
$f^n(x) = e^x(n+x)$.

so it follows that:
$f^n(0) = n$.

and so:
$xe^x = \sum\frac{x^n}{(n-1)!}$ n going from 0 to infinity. n = 1 to infinity

That is:
$xe^x = x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+....$ fine

I am right for the first part?

Second part:

I did the ratio test and I got that the limit is smaller then 1. yes, 0 is < 1
Therefore the sum is absolutely convergent...??? correct

...