Question

find radius of convergence of power series

Answer 1

\[\sum_{n=1}^{\infty} \frac{ n! }{ n^n } x^n\]

Answer 2

You can use the ratio test to do this:
\[\large \begin{align} \lim_{n\rightarrow \infty} \left| \frac{(n+1)!x^{n+1}}{(n+1)^{n+1}}\cdot \frac{n^n}{n!x^n}\right|&= \lim_{n\rightarrow \infty}\left| \frac{(n+1)\cancel{n!x^nx}}{(n+1)^{n+1}}\frac{n^n}{\cancel{n!x^n}}\right|\\
&=\lim_{n\rightarrow \infty}\left|(n+1)^{-n}n^nx \right|\\
\\&=\lim_{n\rightarrow \infty}|x|\left(\frac{n+1}{n}\right)^{-n}\\
&=\lim_{n\rightarrow \infty}|x|\left( 1+\frac{1}{n}\right)^{-n}\\
=&|x|\frac{1}{e}\end{align}\]

Answer 3

You know you'll get convergence when \[ |x|\frac{1}{e}<1\]
Once you simplify that inequality, you should check the endpoints and see if there is convergence at those points by plugging in the value of \(x\) back into your power series.

Answer 4

thankssss !

Answer 5

yw :)