Question

Series question for radius of convergence and interval of convergence: Sum (n=1, infinity) 3(n!)x^n

Answer 1

what do we get as a ratio ?

Answer 2

(n+1)x, with the limit I got x lim (n->infinity) (n+1) which is infinity

Answer 3

3(n+1!) x^(n+1)
---------------
3 (n!) x^n


(n+1)n! x
----------
(n!)


(n+1) x; pull out the x to get:

lim of n+1 as n to inf = inf.

thats the radius i believe

Answer 4

err, sum does not converge that is

Answer 5

Ok, so infinity (-inf, inf)
Let me try it

Answer 6

is this the start of the question? or something that you came to along the way?

Answer 7

"radius of convergence"; this thing doesnt converge

Answer 8

\[\sum_{n=1}^{\infty} 3(n!)x ^{n}\]

Answer 9

That is the question at the start, the actual question

Answer 10

I got that it doesn't converge the first time and it apparently does... somehow

Answer 11

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

example 3 seems to deal with this case

Answer 12

the radius of convergence is 0; but it can still have an interval of convergence, most likely just a single point

Answer 13

hmmm, it says the radius is zero; but the interval are the values of x that make the part pulled out go to zero

Answer 14

|x| = 0 when x=0

Answer 15

... Ok, so whatever makes the x(n+1) go to 0?

Answer 16

Ok, so R = 0 for sure.

Answer 17

yes, when x(inf) = 0

Answer 18

if we consider it this way:

we found\[|x|(\infty)\]therefore
\[|x|(\infty)<1\]
\[|x|<\frac{1}{\infty}\]
\[|x|<0:~:hence~ R=0\]
\[0<x<0:~:interval~=~0\]

Answer 19

interval is: x=0 that is

Answer 20

Thank you! That worked out :)

Answer 21

yay!! :)