Question

Find the radius of convergence for the series from n=1 to infinity of (n^n * x^n) / n!

Answer 1

hows your limit coming along?

Answer 2

I am using the ratio test. I know the taylor series of e^x is series of x^n / n!

Answer 3

The limit I got (n+1)^(n+1) * x / (n+1)

Answer 4

Isn't the numerator always greater so it would be divergent by the divergent test?

Answer 5

\[|x|\lim~\frac{n!~(n+1)^{n+1}}{(n+1)!~n^n}\]
\[|x|\lim~\frac{(n+1)^{n+1}}{(n+1)~n^n}\]
\[|x|\lim~\frac{(n+1)^{n}}{~n^n}\]

Answer 6

We want |n+1/n| to be < 1

Answer 7

not that i see, we just want the limit now

Answer 8

the limit is 1 > 0 which diverges.

Answer 9

but we still have |x| < 1

Answer 10

no, as long as the limit is some finite value, its fine ... then we can work on teh radius of convergence

Answer 11

the radius of convergence is 1.

Answer 12

is that what you are finding?

Answer 13

yes

Answer 14

well, im not finding that

Answer 15

Oh... could you show me what you got?

Answer 16

I thought the limit is x and we want |x|<1 by the ratio test.

Answer 17

i showed you how i got to here
\[|x|\lim~\frac{(n+1)^{n}}{~n^n}\]
\[|x|\lim~\left(\frac{n+1}{n}\right)^n\]
\[|x|\lim~\left(1 + \frac{1}{n}\right)^n\]
now spose the limit of the n parts is some finite value
\[\lim~\left(1 + \frac{1}{n}\right)^n=L\]
then we work the setup:
\[L|x|<1\]
\[|x|<\frac1L\]
and thats our radius

Answer 18

do you know the value of L?

Answer 19

I think it was 1.

Answer 20

its not

Answer 21

lim of (n+1)^n / n^n ? is similar to n^n/ n^n which is 1?

Answer 22

no

Answer 23

\[\lim_{n \to \infty}\left(1+\frac1n\right)^n\implies e \]

Answer 24

How did you get that?

Answer 25

this is something that you prolly should have already been exposed to in a prior problem.

its a bit lengthy to describe and im a bit out of practice.

but if you know anything about a compounding rate of interest....

Answer 26

Also, is there any way we could solve this problem by using this info: the taylor series of e^x is series of x^n / n!

Answer 27

Thank you so much btw :)

Answer 28

im not sure if we can or not, the radius of convergence is not dependant on a 'parent' function

f(x) e^x does not necessarily have the same radius of convergence as e^x

Answer 29

Ok

Answer 30

i wish i could recall a good way to prove the limits e, but nothings coming to mind :) been too long

Answer 31

Alrighty, thank you for your help anyway :) I really appreciate it

Answer 32

L'Hopital's rule will do the trick

Answer 33

ohhhhh ok

Answer 34

Provided you write the limit in indeterminate form, of course

Answer 35

sounds good, thank you:)