Question

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 ):

Answer 1

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

Answer 2

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?

Answer 3

YES. and then i added 2 to both sides. so 0<x<4 ?

Answer 4

that is what i got but i kind of guessed. ok so now we have to check at x = 0 and x = 4 right?

Answer 5

yerp....

Answer 6

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

Answer 7

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

Answer 8

i cancled the
\[2^n\]

Answer 9

oooh lol sorry

Answer 10

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

Answer 11

i canceled wrong...

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

kick the minus sign in to the first part to make it

\[(-1)^n\] instead of
\[(-1)^{n+1}\]

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

haha okay.

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

wait wait!!

Answer 24

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

Answer 25

why? i think it is correct...

Answer 26

let me be more careful with zero

Answer 27

whoa did i make a mistake!! lets go slow

Answer 28

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

Answer 29

lets look at the numerator when x = 0

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

ohh i seee...

Answer 34

ratio test?

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

oh okay i get it!!

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

thanks !

Answer 48

yw