Question

prove that the sequence sin(n*pi/2) diverges using the form (some epsilon)(for all N)(some n)(n>N => |x_n - L| >= epsilon)

Answer 1

to prove the divergence ! change "=>" by "and"

Answer 2

right ty

Answer 3

what you typed is correct ! just change "=>" by and

Answer 4

ohh ! I didn't see your reply ! sorry !

Answer 5

I know for something like (-1)^n we make a case for n is odd so x_n is -1 and let L >0
and case 2 we let n be even and L<=0. And then show that (x_n-L)>= 1 for some n > all N this is true. I guess I dont understand why we led n odd go with L<0 and n even go with L>= 0.

Answer 6

Is that a solution ?! because if you're proving the divergence you don't have the right to choose L anymore ! (cause implicitly there is for all L in R before some epsilon !)

Answer 7

well it did two cases one for L>=0 and L<0, is that choosing? Im sorry im very new to this stuff and this is my first upper dev class.

Answer 8

now ! I understand ! just begin with L ! not n ....the cases are for L !
and no problem ! I'm trying to make you understand ! ( well I'm a bad teacher...)

Answer 9

yeah I just dont understand why we did not have to show both cases of n odd and both cases for n even.

Answer 10

you don't have to !
let L>0 so we "have to choose" epsilon=.....as 1
let N be any positive integer !( now we have to choose n) let's take n= 2(N)+1
so obviously n>N |x_n - L|= |-1 - L|=|1 +L| (-1 is a trick to take off | | and...)
1+L>0 then | 1+L|=1+L >=1
the same for L<=0
so n is choosen (using N if we want to !)

Answer 11

sorry ! Be right back !

Answer 12

ahhh I c, we get to choose epsilon and n

Answer 13

yes ! :) brb

Answer 14

I'm Back !

Answer 15

word.
so I can have a case for l > 0 and n = pi/2 + (n-1)(2pi)?

Answer 16

and epsilon = 1/2

Answer 17

for your exercise ?!

Answer 18

yeah the sin(npi/2)

Answer 19

There are many many many ways to do this ! You can do it without using cases !

Answer 20

well thats the only way I can think to do it. but I dont know what dto do for the the case when sin(npi/2) = 0

Answer 21

why are you making cases fo "n" ? you don't have to !

Answer 22

you can choose epsilon using L....But you have to force it to differ from zero and also satisfies 0< epsilon

Answer 23

can you show me?

Answer 24

Just do it with cases if you can !

Answer 25

Sorry ! I made a mistake ! With cases is easier !

Answer 26

Sorry Again ! Just do the same thing As you've done for (-1)^n
\[if L >0 \]
then let epsilon =1/2 or (1=)
let N be any positive integer ! let n= 4(N+1)+3 then sin(n pi/2)=-1
| -1-L|=|1+l=1+L>1/2
foe L<=0.....the same
just n = 4(N+1)+1 then |sin( npi/2)-L|=|1-L|=1-L>1/2 (becaus -L>=0)

sorry for confusing you
@zzr0ck3r

Answer 27

nope hah great that is almost exactly what i did just a variation on the n

Answer 28

Great ! you're learning too fast :) !

Answer 29

thank you man, you opened a door for me today and last night. Been waiting for that door to open for about 24 hours:)

Answer 30

you're welcome ! :) !