Question

Can someone assist me with this question please. Determine the radius of convergence of the power series sum of n from zero to infinite of (n!x^n)/n^n

Answer 1

what is the ratio n+1/n

Answer 2

1

Answer 3

\[\lim\frac{(n+1)~x^{n+1}}{(n+1)^{n+1}}\frac{n^n}{n!~x^n}\]
\[\lim\frac{(n+1)~x}{(n+1)^{n+1}}\frac{n^n}{1}\]
\[|x|\lim\frac{(n+1)~n^n}{(n+1)^{n+1}}\]
\[|x|\lim\frac{n^{n+1}+....}{n^{n+1}+.....}=|x|\]
\[|x|<1~:~radius~=1\]
\[interval:~-1<x<1\]

Answer 4

i didnt mess that up did i?

Answer 5

http://www.wolframalpha.com/input/?i=sum+%28n%21x%5En%2Fn%5En%29

the wolf suggests that the radius is e, if the index starts at 1

Answer 6

(n!x^n)/n^n
$$C=\lim_{n\to\infty}\left|\frac{n^n(n+1)!\ x^{n+1}}{(n+1)^{n+1}\,n!\,x^n}\right|=\lim_{n\to\infty}\left|\frac{n^n(n+1)x}{(n+1)^{n+1}}\right|=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}|x|<1\\\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n=\left(1-\frac1{n+1}\right)^n\sim\left(1-\frac1n\right)^n=\frac1e\text{ as }n\to\infty\\\text{so }C=\frac1e|x|<1\implies |x|<e$$