Question

The interval of convergence of the series
[(3^n(x-2)^n)/ (n!)] is from n=0 to infinity.

Answer 1

\[ \sum_{n=0}^{\infty} [(3^n(x-2)^n/ n!)]\]

Answer 2

To be clear numerator: 3^n(x-2)^n
Denominator: n!

Answer 3

nice :)

Answer 4

im so bad with series questions...i dont even know where to start with some of them

Answer 5

i recall this vaguely, so ill have to resort to pauls workings of it

Answer 6

ok

Answer 7

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

if you wanna make sure i read it right ;)

Answer 8

it looks like we take the ratio like in a ratio test

Answer 9

yeah i rmbr my teaching saying if the series has a factorial and a power u use the ratio test

Answer 10

\[\sum_{n=0}^{\infty} \frac{3^n(x-2)^n}{n!}\]
\[\lim_{n\to inf} \frac{3^n(x-2)^n}{n!} \frac{(n-1)!}{3^{(n-1)}(x-2)^{(n-1)}}\]

\[\lim_{n\to inf} \frac{3(x-2)}{n}\]

Answer 11

as n get large; the value goes to zero

Answer 12

i have a qustion

Answer 13

which I believe means that it is convergent everywhere

Answer 14

i thought for the ratio test ur supppsed to add 1 to n not subtract

Answer 15

so like it shud be 3^(n+1)* (x-2)^n+1

Answer 16

the ration test is of the form\[\frac{A_{now}}{A_{past}}\]or\[\frac{A_{future}}{A_{now}}\]

Answer 17

the idea being that your ratio is a proportion of 2 terms with the top being the one right after the bottom

Answer 18

so either n+1 the left part; or n-1 the right part

Answer 19

wat do u mean left and right part

Answer 20

srry my teacher never talked about ratio test like that

Answer 21

\[\lim_{n\to inf} \frac{3^{(n+1)}(x-2)^{(n+1)}}{(n+1)!} \frac{n!}{3^{n}(x-2)^{n}}\]

\[\lim_{n\to inf} \frac{3(x-2)}{n+1}\]

same results over all

Answer 22

since our ratio is a fraction over a fraction .....
\[\cfrac{\frac{a}{b}}{\frac{c}{d}}\to\cfrac{\frac{a}{b}}{\frac{c}{d}}*\cfrac{\frac{d}{c}}{\frac{d}{c}}= \frac{a}{b}\frac{d}{c}\]

Answer 23

we end up with a left and a right part to this

Answer 24

so now how do we find the interval of converegence since that is wat the question is asking

Answer 25

well, we view the results of our limit ratio

Answer 26

its zero; which tells us that the interval on which this converges is from -inf to inf

Answer 27

example 4 of the link i posted gives a reasoning as to why i believe :)

Answer 28

oh i see. thanks for helping. =)

Answer 29

youre welcome