Question

Power Series: sigma ((-1)^n x^n)/(n+1) as n-> infinity

Answer 1

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

Answer 2

i know that im suppose to replace all n's with n+1 and then use ration test but im not sure what steps to take next

Answer 3

Hi donutboy, how are you? Do you know the series representation for ln(1-x)?

Answer 4

You want to check the convergence of the series or you want to find its sum?

Answer 5

ratio* hi sklee, im fine, no i dont know the ln(1-x)

Answer 6

im looking for the radius of convergence and interval of convergence

Answer 7

ok great, if you want to find the radius and interval of convergence, then you dont have to know that series representation

Answer 8

For radius of convergence, i assume you will use the Ration test, and not using the formula. So I will show you the work using the test.

Answer 9

Let a_n = (-1)^n x^n / (n+1), then by the Ratio test,

lim |a_(n+1)/a_n| = lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)]
= lim |x {(n+1)/(n+2)}|

Because lim (n+1)/(n+2) = 1 and the series converges if the value of the limit of |a_(n+1)/a_n| is less than 1, we get

|x|<1

Answer 10

ahh icic thank you so much! my prof. goes way to fast and its hard to keep up

Answer 11

So the radius of convergence is 1, and the interval we are considering is -1 < x < 1. So we need to check the convergence of the series at the end point x =-1 and x =1

Answer 12

at x = -1, the series become \sum {1/(n+1)}, which is divergent as it is a variation of harmonic series.

at x=1, the series becomes \sum {(-1)^n / (n+1)}, which is a variation of the Alternating harmonic series, and this series converges.

So the interval of convergence is -1 < x =< 1

Answer 13

quesiton on the lim |a_(n+1)/a_n| = lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)]
= lim |x {(n+1)/(n+2)}| part what happens during lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)] ?

Answer 14

x^(n+1)/x^n can be simplified to become x

Answer 15

and i just arrange the term (n+1) and (n+2) as a fraction

Answer 16

oh okay thank you again

Answer 17

you're welcome

Answer 18

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

Answer 19

so far ive gotten to \[\lim_{n \rightarrow \infty} \left| (2n!(x^(2n+1)/x^(2n)(2n+2) \right|\]

Answer 20

ok, very good

Answer 21

now the [x^(2n+1)]/x^(2n) is simplified to x

Answer 22

and (2n)! / (2n+2)! is simplified to 1/[(2n+1)(2n+2)] because (2n+2)! = (2n+2)(2n+1)[(2n)!]

Answer 23

alrighty

Answer 24

the limit of 1/[(2n+1)(2n+2)] is 0, so the series converges for all x

Answer 25

so the radius is infinity and the interval of convergence is (-infinity, infinity)

Answer 26

yup, you are right!

Answer 27

thanks! im starting to get this now!

Answer 28

great, you can do it donutboy. All the best!

Answer 29

thanks ^__^ youre awesome!