anonymous
  • anonymous
i need to find the interval of convergence of the series (from 1 to infinity) of ((-1^n+1)(x-2)^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 ):
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sum_1^{\infty}\frac{(-1)^{n+1}(x-2)^n}{n2^n}\]
anonymous
  • anonymous
i am going to make a guess that the radius is 2, so that you will have \[|x-2|<2\] is that what you got?
anonymous
  • anonymous
YES. and then i added 2 to both sides. so 0

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anonymous
  • anonymous
that is what i got but i kind of guessed. ok so now we have to check at x = 0 and x = 4 right?
anonymous
  • anonymous
yerp....
anonymous
  • anonymous
maybe 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
  • anonymous
wait. how did you do that. how does it cancel off?
anonymous
  • anonymous
i cancled the \[2^n\]
anonymous
  • anonymous
oooh lol sorry
anonymous
  • anonymous
my mistake i meant \[\sum\frac{(-1)^n}{n}\] oops
anonymous
  • anonymous
i canceled wrong...
anonymous
  • anonymous
no actually my algebra is terrible, can you explain the cancelling please?
anonymous
  • anonymous
but conclusion is right. converges because you have an alternating series wehre terms go to zero
anonymous
  • anonymous
you 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
  • anonymous
kick the minus sign in to the first part to make it \[(-1)^n\] instead of \[(-1)^{n+1}\]
anonymous
  • anonymous
that just leaves you with \[\sum\frac{(-1)^n}{n}\]
anonymous
  • anonymous
ohh okay. and then by alternating series test it converges. got it! what about at x=4?
anonymous
  • anonymous
well i didn't do that yet, so lets try it
anonymous
  • anonymous
haha okay.
anonymous
  • anonymous
i think this one is a problem because it will not alternate
anonymous
  • anonymous
i think it will alternate? and it diverges because 1/n is harmonic.
anonymous
  • anonymous
thank you so much though! do you know how to differentiate and integrate the series ones too?
anonymous
  • anonymous
wait wait!!
anonymous
  • anonymous
hold on a second i think i might have made a big big mistake!!
anonymous
  • anonymous
why? i think it is correct...
anonymous
  • anonymous
let me be more careful with zero
anonymous
  • anonymous
whoa did i make a mistake!! lets go slow
anonymous
  • anonymous
oh oops i think the x=4 is convergent too.
anonymous
  • anonymous
lets look at the numerator when x = 0
anonymous
  • anonymous
\[(-1)^{n+1}(-2)^n\] \[(-1)^{n+1}(-1)^n2^n\] \[(-1)^{2n+1}2^n\]
anonymous
  • anonymous
now we can certainly cancel a \[2^n\] with the denominator
anonymous
  • anonymous
get \[\sum\frac{(-1)^{2n+1}}{n}\]
anonymous
  • anonymous
ohh i seee...
anonymous
  • anonymous
ratio test?
anonymous
  • anonymous
but \[2n+1\] is odd, and so the numerator is -1
anonymous
  • anonymous
in other words you get \[\sum\frac{-1}{n}\] which diverges for sure
anonymous
  • anonymous
ohhh okay okay. sounds good. um can you help me with another one?
anonymous
  • anonymous
sorry 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
  • anonymous
we did not do it with 4 yet, wanna try it?
anonymous
  • anonymous
4 is going to converge, because this time the numerator is \[(-1)^{n+1}2^n\]
anonymous
  • anonymous
cancel the \[2^n\] get \[\sum \frac{(-1)^{n+1}}{n}\]
anonymous
  • anonymous
and this one alternates, so it converges for x = 4 but not for x = 0
anonymous
  • anonymous
oh okay i get it!!
anonymous
  • anonymous
yeah well i made a mistake, sorry it took so long
anonymous
  • anonymous
you were so much help regardless. i barely know what i am doing. can i ask you for help on another one?
anonymous
  • anonymous
sure but post in a new thread and i will come look it is much easier than scrolling down
anonymous
  • anonymous
thanks !
anonymous
  • anonymous
yw

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