Question

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Answer 1

so this is like \[\sum_{n=0}^{\infty}\left( \frac{ x }{ 3 } \right)^n\]

this is a geometric series so \(|\frac{x}{3}|<1\)

Answer 2

ooops, yours starts at n=1. no biggie, same must be true.

Answer 3

Could I also somehow find the answer using the ratio test? I just don't really understand how that is the answer

Answer 4

it's not the answer... you have to expand.

\(-1<\frac{x}{3}<1\) solve for x. you know about geometric series, right? the ratio, r, must be between -1 and 1.

Answer 5

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

Answer 6

To use the ratio test compare the ratio of the (n+1)st term to the nth term. If the ratio
has absolute value less than 1 for large n, the series converges. In your case, the (n+1)st term is \[(x/3)^{(n+1)}\] and the nth term is \{(x/3)^n\], so you need

\[\vert (x/3)^{(n+1)}/(x/3)^n\vert < 1\]

or \[ \vert x/3\vert <1\]

Answer 7

hint:

multiply all part by 3 \(-1<\frac{x}{3}<1\) to find the radius of convergence.

Answer 8

I know, that's what I did

Answer 9

okay, sorry

Answer 10

and I got x>-3 and x<3

Answer 11

no... -3<x<3 it's bounded

Answer 12

yeah, sorry, same thing

Answer 13

Is the interval of convergence the same as the radius of convergence?

Answer 14

oh, sorry, that would be the interval of convergence.

the radius of convergence should just be 0<r<3

Answer 15

i think the 0 part is less that or equal to.

Answer 16

i got asked the interval of convergence so yeah, it's fine :) I was just wondering

Answer 17

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

hav a look at this. it explains fairly well what's going on. the trick with the radius/interval thing has to do with the endpoints of the interval and whether or not the series is convergent or not at the endpoints.