Question

The radius of convergence for the power series (x-3)^2n/n from n=1 to infinity is equal to 1. What is the interval of convergence?

Answer 1

what do you think it is?

Answer 2

well i thought it was from -1 to 1 but its wrong. I know the anwer but just not how to get it

Answer 3

it has to be centered at 3

Answer 4

I was trying to use the ratio test but i guess i did something wrong

Answer 5

not zero

Answer 6

why 3?

Answer 7

with the ratio test you will end up with

|x-3|<1

Answer 8

can pleas show me how 2 do the ratio test. i can usually do the probem from there. it ust teh test i have trouble with

Answer 9

hold on....is this your series...

\[\sum_{n=1}^{\infty}\frac{(x-3)^{2n}}{n}\]

Answer 10

yes andthats equal to 1

Answer 11

ok...so you still get |x-3|<1

\[\Rightarrow 2<x<4\]

Answer 12

the radius of convergence is 1...not the sum

Answer 13

oh right...sorry. Can u show me how u got x-3 less than 1

Answer 14

ratio test...

Answer 15

let \[a_n=\frac{(x-3)^{2n}}{n}\]

then \[\left|\frac{a_{n+1}}{a_n}\right|\]

\[=\left|\frac{\frac{(x-3)^{2(n+1)}}{n+1}}{\frac{(x-3)^{2n}}{n}}\right|\]


simplify this and take the limit as \(n\to \infty\)

Answer 16

i understand how to get that but i dont ever get the simplying part right. im not sure how to cancel out things.

Answer 17

ratio test says that you will have convergence if the above limit is less than 1

Answer 18

\[=\left|\frac{\frac{(x-3)^{2(n+1)}}{n+1}}{\frac{(x-3)^{2n}}{n}}\right|\]

\[=\left|\frac{(x-3)^{2(n+1)}}{n+1} \frac{n}{(x-3)^{2n}}\right|\]

\[=\left|\frac{(x-3)^{2n+2}}{n+1} \frac{n}{(x-3)^{2n}}\right|\]

\[=\left|\frac{(x-3)^{2}}{n+1} n\right|\]

\[=\left|(x-3)^{2}\frac{n}{n+1} \right|\to|x-3|^2\text{ as }n\to\infty\]

Answer 19

then \[|x-3|^2<1\Rightarrow |x-3|<1\]

Answer 20

right so then u add 3 and u get x betwen 2 and 4. so now we have to test them to see which is included

Answer 21

yes...you now need to test the boundary

Answer 22

you sould be careful when finding the interval

\[|x-3|<1\]

\[X-3<1\]
and \[-(x-3)<1\]



first one

\[X-3<1\Rightarrow x<1+3=4\]

for the second

\[-(x-3)<1\Rightarrow x-3>-1\Rightarrow x>3-1=2\]



thus \[2<x<4\]

Answer 23

so b/c its equal to a number greater than 1 it diverges

Answer 24

sorry but i dont understand the difference btwn X and x

Answer 25

plug 2 and 4 gack into the original series for x and see if they converge

Answer 26

just a typo...should all be x

Answer 27

oh ok i see....THANKS SOO MUCH!

Answer 28

good