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 ):

- anonymous

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- anonymous

\[\sum_1^{\infty}\frac{(-1)^{n+1}(x-2)^n}{n2^n}\]

- 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

YES. and then i added 2 to both sides. so 0

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## More answers

- 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

yerp....

- 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

wait. how did you do that. how does it cancel off?

- anonymous

i cancled the
\[2^n\]

- anonymous

oooh lol sorry

- anonymous

my mistake i meant
\[\sum\frac{(-1)^n}{n}\] oops

- anonymous

i canceled wrong...

- anonymous

no actually my algebra is terrible, can you explain the cancelling please?

- anonymous

but conclusion is right. converges because you have an alternating series wehre terms go to zero

- 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

kick the minus sign in to the first part to make it
\[(-1)^n\] instead of
\[(-1)^{n+1}\]

- anonymous

that just leaves you with
\[\sum\frac{(-1)^n}{n}\]

- anonymous

ohh okay. and then by alternating series test it converges. got it! what about at x=4?

- anonymous

well i didn't do that yet, so lets try it

- anonymous

haha okay.

- anonymous

i think this one is a problem because it will not alternate

- anonymous

i think it will alternate? and it diverges because 1/n is harmonic.

- anonymous

thank you so much though! do you know how to differentiate and integrate the series ones too?

- anonymous

wait wait!!

- anonymous

hold on a second i think i might have made a big big mistake!!

- anonymous

why? i think it is correct...

- anonymous

let me be more careful with zero

- anonymous

whoa did i make a mistake!! lets go slow

- anonymous

oh oops i think the x=4 is convergent too.

- anonymous

lets look at the numerator when x = 0

- anonymous

\[(-1)^{n+1}(-2)^n\]
\[(-1)^{n+1}(-1)^n2^n\]
\[(-1)^{2n+1}2^n\]

- anonymous

now we can certainly cancel a
\[2^n\] with the denominator

- anonymous

get
\[\sum\frac{(-1)^{2n+1}}{n}\]

- anonymous

ohh i seee...

- anonymous

ratio test?

- anonymous

but
\[2n+1\] is odd, and so the numerator is -1

- anonymous

in other words you get
\[\sum\frac{-1}{n}\] which diverges for sure

- anonymous

ohhh okay okay. sounds good. um can you help me with another one?

- 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

we did not do it with 4 yet, wanna try it?

- anonymous

4 is going to converge, because this time the numerator is
\[(-1)^{n+1}2^n\]

- anonymous

cancel the
\[2^n\] get
\[\sum \frac{(-1)^{n+1}}{n}\]

- anonymous

and this one alternates, so it converges for x = 4 but not for x = 0

- anonymous

oh okay i get it!!

- anonymous

yeah well i made a mistake, sorry it took so long

- anonymous

you were so much help regardless. i barely know what i am doing. can i ask you for help on another one?

- anonymous

sure but post in a new thread and i will come look
it is much easier than scrolling down

- anonymous

thanks !

- anonymous

yw

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