Question

Find the series interval of convergence and within this interval the sum of the series as a function of x

Answer 1

\[\sum_{0}^{\infty} (x+1)^{2n}/9^{n}\]

Answer 2

Can I integrate? hehe

Answer 3

this is continuous function right?

Answer 4

but i think it would be a waste to integrate

Answer 5

how to do write the "over" division line without it including summation

Answer 6

Hmm \frac{}{} <<use this

Answer 7

when you said integrate.. what did you mean?

Answer 8

Hmm that was kinds foolish I guess umm I was talking about integrating the function from o to infinity but i don't think it's integration is possible

Answer 9

is that something to do with taylor series? because i have not learned that yet.

Answer 10

taylor series hmm it's about differentiating the function about zero hmm i have a good video link for it lemme fetch that for you

Answer 11

http://www.khanacademy.org/video/taylor-series-at-0--maclaurin--for-e-to-the-x?playlist=Calculus
try this :-D i learned about taylor series from khan academy only but somehow i don't remember much of it, poor memory :/

Answer 12

you don't have to taylor series. It's the root test first to get the interval of convergence.

Answer 13

\[\sum_{0}^{\infty} \frac{}{}\ \left( (x+1)^2\over9 \right)^n\]

Answer 14

?

Answer 15

|r| <1 could i just solve for the intervals of convergence?

Answer 16

soo..

∑0∞(x+1)2n/9n = \[\sum_{1}^{\infty}\left| (an+1) \div an \right| = L < 1\]

Answer 17

-4<x<2 abs convergent because it would be geometric progression,
i thought the root test was |an|^1/n

Answer 18

ok then you test the two endpoints. x=-4 x=2 for absolute convergence

Answer 19

is that okay? im still unsure of myself with power series

Answer 20

and to sum the series as a function of x, just use a1/1-r? where r = (1/9)(x-1)^2

Answer 21

well thanks for you help franklin

Answer 22

its getting hard to focus.. thanks for your help ishaan

Answer 23

yea sorry I was trying to find my old notes, forgot how to rewrite the sumation

Answer 24

sleep is calling

Answer 25

good luck

Answer 26

thanks!

Answer 27

stupid even numbered problems :)