Question

Power series

Answer 1

\[\sum_{n=0}^{\infty} n!x^n\]

Answer 2

@SithsAndGiggles

Answer 3

If I use the ratio test, I would eventually get \[\lim_{n \rightarrow \infty} (n+1)x\]

but why do we assume x = 0?

Answer 4

\[\lim_{n\to\infty}\left|\frac{(n+1)!x^{n+1}}{n!x^n}\right|=\lim_{n\to\infty}\left|(n+1)x^n\right|=|x|\lim_{n\to\infty}(n+1)\]
The limit is not finite for any non-zero value of \(x\). For the ratio test to work to establish convergence of the series, the limit must be less than 1.

Answer 5

This means the series only converges for one point, \(x=0\), and diverges elsewhere.

Answer 6

However, given that the series starts at \(n=0\), we can't say it converges for \(x=0\) because \(0^0\) in an indeterminate form.

Answer 7

Gotcha, thank you very much!

Answer 8

You're welcome!