- anonymous

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

- chestercat

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- anonymous

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

- anonymous

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

- anonymous

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

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- anonymous

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

- anonymous

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

- anonymous

im looking for the radius of convergence and interval of convergence

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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)] ?

- anonymous

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

- anonymous

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

- anonymous

oh okay thank you again

- anonymous

you're welcome

- anonymous

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

- anonymous

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

- anonymous

ok, very good

- anonymous

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

- anonymous

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

- anonymous

alrighty

- anonymous

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

- anonymous

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

- anonymous

yup, you are right!

- anonymous

thanks! im starting to get this now!

- anonymous

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

- anonymous

thanks ^__^ youre awesome!

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