Question

Use the ratio test to find the radius of convergence and interval of convergence of the power series.

Answer 1

\[x-(x^2/4)+(x^3/9)-(x^4/16)+(x^5/25)-...\]

Answer 2

this the given that you write? x−(x2/4)+(x3/9)−(x4/16)+(x5/25)−...

Answer 3

yes this is the series

Answer 4

how do i find the radius of convergence of this power series ?

Answer 5

http://answers.yahoo.com/question/index?qid=20110407230807AA57QNX

Answer 6

you are not giving the answer this different series....

Answer 7

what?

Answer 8

Do you know how to use the Ratio Test?

Answer 9

math give a link that not helpful for you he giving you answer for different series

Answer 10

@nader1 im helping with similar problem

Answer 11

yes i know how to use the ratio test

Answer 12

how do i find the sigma notation of this series?

Answer 13

Say you have a series
\[\huge \sum_{n=1}^\infty a_n\]Then if you consider the limit
\[\huge k=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|\]
Then if
k > 0, then it's divergent
k < 0, then it's absolutely convergent
k = 1, no conclusion may be drawn

Answer 14

Now as for a sigma notation, look for a pattern... all power series take the form
\[\huge \sum_{n=1}^\infty a_nx^n\]right?

Answer 15

So, let's examine your series...
\[\Large x -\frac{x^2}{4}+\frac{x^3}{9}-\frac{x^4}{16}+\frac{x^5}{25}-...\]
see a pattern?

Answer 16

x^n/n^2?

Answer 17

Not quite :)
As you can see, the signs alternate.

Answer 18

\[(-1)^{n-1} \frac{ x^n }{ n^2 }\]

Answer 19

?

Answer 20

Yeah, that's it :)
\[\huge \sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n^2}\]

Answer 21

Not that the -1 really matters, you're going to take the absolute value anyway, when you use the ratio test. Please do that now ^.^

Answer 22

i do the ratio test now?

Answer 23

Yes.

Answer 24

ugh is a long one

Answer 25

is there a trick to do it faster

Answer 26

Not really :)
After all, you're getting absolute value, means that that
\[(-1)^{n+1}\]won't mean squat when you're using the Ratio test.

Answer 27

So, using the ratio test, we need this limit
\[\huge k = \lim_{n\rightarrow\infty}\left|\frac{\frac{x^{n+1}}{(n+1)^2}}{\frac{x^n}{n^2}}\right|\]

Answer 28

can you reduce it to \[\frac{ x*(n+1)^2 }{ n^2 }\]

Answer 29

That you can :)

Answer 30

Don't forget absolute value, ok?

Answer 31

so limit is x?

Answer 32

Almost, actually, it's |x|

Answer 33

oh yea

Answer 34

So, can you do it from here?

Answer 35

how do i find the radius from here?

Answer 36

The limit k in your application of the Ratio Test was |x| right?
Well, what must that limit be, if your series is to be convergent?

Answer 37

1?

Answer 38

That's right, or less :)
So,
\[\large |x|\le1\]There's your radius of convergence

Answer 39

Whoops, sorry
\[\large |x| < 1\]

Answer 40

thank you

Answer 41

No problem :)