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anonymous
 4 years ago
i need to find the interval of convergence of the series (from 1 to infinity) of ((1^n+1)(x2)^n)/n2^n and then check for the convergence at the endpoints. i got the interval but i am having trouble with checking the endpoints. help ):
anonymous
 4 years ago
i need to find the interval of convergence of the series (from 1 to infinity) of ((1^n+1)(x2)^n)/n2^n and then check for the convergence at the endpoints. i got the interval but i am having trouble with checking the endpoints. help ):

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_1^{\infty}\frac{(1)^{n+1}(x2)^n}{n2^n}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am going to make a guess that the radius is 2, so that you will have \[x2<2\] is that what you got?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0YES. and then i added 2 to both sides. so 0<x<4 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is what i got but i kind of guessed. ok so now we have to check at x = 0 and x = 4 right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe i am off by a minus sign, not sure, but this is an alternating series so surely converges as the terms go to zero, so we can include 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait. how did you do that. how does it cancel off?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i cancled the \[2^n\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my mistake i meant \[\sum\frac{(1)^n}{n}\] oops

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no actually my algebra is terrible, can you explain the cancelling please?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but conclusion is right. converges because you have an alternating series wehre terms go to zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you have in the numerator \[(2)^n\] and in the denominator you have \[2^n\] but \[(2)^n=(1)^n2^n\] so you can cancel a \[2^n\]top and bottom

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0kick the minus sign in to the first part to make it \[(1)^n\] instead of \[(1)^{n+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that just leaves you with \[\sum\frac{(1)^n}{n}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh okay. and then by alternating series test it converges. got it! what about at x=4?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well i didn't do that yet, so lets try it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think this one is a problem because it will not alternate

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think it will alternate? and it diverges because 1/n is harmonic.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you so much though! do you know how to differentiate and integrate the series ones too?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hold on a second i think i might have made a big big mistake!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why? i think it is correct...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me be more careful with zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whoa did i make a mistake!! lets go slow

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh oops i think the x=4 is convergent too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lets look at the numerator when x = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(1)^{n+1}(2)^n\] \[(1)^{n+1}(1)^n2^n\] \[(1)^{2n+1}2^n\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now we can certainly cancel a \[2^n\] with the denominator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0get \[\sum\frac{(1)^{2n+1}}{n}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but \[2n+1\] is odd, and so the numerator is 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in other words you get \[\sum\frac{1}{n}\] which diverges for sure

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh okay okay. sounds good. um can you help me with another one?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry about that , it was a big mistake on my part. i just thought to kick over the minus sign, but i forgot to add the exponents

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we did not do it with 4 yet, wanna try it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.04 is going to converge, because this time the numerator is \[(1)^{n+1}2^n\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cancel the \[2^n\] get \[\sum \frac{(1)^{n+1}}{n}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and this one alternates, so it converges for x = 4 but not for x = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah well i made a mistake, sorry it took so long

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you were so much help regardless. i barely know what i am doing. can i ask you for help on another one?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure but post in a new thread and i will come look it is much easier than scrolling down
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