Question

convergence test!
\[\sum_{1}^{\infty}\frac{n!e^n}{n^n}\]
we are doing in the class power series and trying to test does the series \[\sum_{1}^{\infty}\frac{n!x^{2n}}{n^{n}} ~~for~~ x=\sqrt{e}\] converge or not

Answer 1

i tried to show that the limit of the nth term is not 0 therefore the series must diverge
seems something is not good

Answer 2

everything looks good, how did you compute the limit ?

Answer 3

hmm this is what i did \[\lim_{n\to \infty }n!(\frac{e}{n})^n=L, ~~L\ne0\]
applied log to both sides
but seems i have to difficulty to go further lol

Answer 4

try using stirling approximation
http://mathworld.wolfram.com/StirlingsApproximation.html

Answer 5

hmm ok let me see

Answer 6

\[\lim n\ln-n+n-n\ln n=\ln L \Longrightarrow L=1 \]

Answer 7

Am i right? using that approach lnn! is nlnn-n

Answer 8

seems that the limit on the left side is 0

Answer 9

let me get paper and pen

Answer 10

alright :)

Answer 11

we may use this directly instead
http://gyazo.com/8165782b98540ff87489e00fa793bf97

Answer 12

replace \(n!\) by \(n^ne^{-n}\sqrt{2\pi n}\)

Answer 13

\[\lim\limits_{n\to\infty}\frac{\color{blue}{n!}e^n}{n^n}=\lim\limits_{n\to\infty}\frac{\color{blue}{n^ne^{-n}\sqrt{2\pi n}}e^n}{n^n}\]

simplify and evaluate

Answer 14

that goes to infinity which is good

Answer 15

thanks :)