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anonymous
 5 years ago
Power Series: sigma ((1)^n x^n)/(n+1) as n> infinity
anonymous
 5 years ago
Power Series: sigma ((1)^n x^n)/(n+1) as n> infinity

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{0}^{\infty} ((1)^n x^n)/(n+1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Hi donutboy, how are you? Do you know the series representation for ln(1x)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You want to check the convergence of the series or you want to find its sum?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ratio* hi sklee, im fine, no i dont know the ln(1x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im looking for the radius of convergence and interval of convergence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok great, if you want to find the radius and interval of convergence, then you dont have to know that series representation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For 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
 5 years ago
Best ResponseYou've already chosen the best response.0Let 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
 5 years ago
Best ResponseYou've already chosen the best response.0ahh icic thank you so much! my prof. goes way to fast and its hard to keep up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0at 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
 5 years ago
Best ResponseYou've already chosen the best response.0quesiton 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
 5 years ago
Best ResponseYou've already chosen the best response.0x^(n+1)/x^n can be simplified to become x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i just arrange the term (n+1) and (n+2) as a fraction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh okay thank you again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{0}^{\infty} (1)^n (x^(2n)/ (2n!)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so far ive gotten to \[\lim_{n \rightarrow \infty} \left (2n!(x^(2n+1)/x^(2n)(2n+2) \right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now the [x^(2n+1)]/x^(2n) is simplified to x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and (2n)! / (2n+2)! is simplified to 1/[(2n+1)(2n+2)] because (2n+2)! = (2n+2)(2n+1)[(2n)!]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the limit of 1/[(2n+1)(2n+2)] is 0, so the series converges for all x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the radius is infinity and the interval of convergence is (infinity, infinity)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks! im starting to get this now!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great, you can do it donutboy. All the best!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks ^__^ youre awesome!
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