Question

prove that the sequence (-9/10)^n converges to 0

Answer 1

I think you have to use the alternating series test.

Answer 2

this is a sequence not a series

Answer 3

no sumations

Answer 4

it should have hte form \[|x_{n}-L| < epsilon\]

Answer 5

bah equation editor is being lame

Answer 6

oh, wait. yeah. sorry, i read it wrong.

so you take the lim as n-> infinity of |r|^n.

so lim n-> infinity of |9/10|^n = 0.

so by the monotone convergence thm, it is convergent.

also one of the properties of the limits of sequences is that if the limit of the absolute value of the sequence goes to 0, then so does the original series as well.

Answer 7

I have to prove is using the form above

Answer 8

hmm. maybe @TuringTest could help with that. I don't quite know how to put it in that form.

Answer 9

did you end up getting it? :S

Answer 10

the way i would show it converges, is prove the limit as n->infinity is zero
\[\lim_{n \rightarrow \infty}(-\frac{9}{10})^{n} = \lim_{n \rightarrow \infty}\frac{(-1)^{n}}{(\frac{10}{9})^{n}} = 0\]

Answer 11

sorry im not great at using epsilon delta proofs for these type of problems

Answer 12

let
\[\epsilon >0\]
let \[N=\max( \lfloor \frac{-\ln(\epsilon)}{\ln(\frac{10}{9})} \rfloor+1;1)\]
and let n > N
then we will have \[\frac{-\ln( \epsilon)}{\ln(\frac{9}{10})} <N<n\]
so , \[-\ln(\epsilon )<nln(\frac{10}{9})\]
then \[\ln( \epsilon )>-nln(\frac{10}{9})=nln(\frac{9}{10})=\ln(\left|{(\frac{ -9}{10}})^n \right|)\]
then wich mean that \[\epsilon>\left|{(\frac{ -9}{10}})^n \right|\]
this finish the proof !

Answer 13

ok thats exactly what I did, but didnt know if I was missing something tricky with ln. ty

Answer 14

you're welcome ! :)

Answer 15

so I dont know what the max thing is but I said let N > ln(1/ep)/ln(10/9) and n>N

Answer 16

max is the greater number of both numbers !
and be careful ! you have to choose N as a positive integer !

Answer 17

yeah, hmm I still dont understand why you used the max thing. We have not done anything like this so I just may need to think about it for a bit.

Answer 18

just to make sure that my N is always a positive integer ! if epsilon >1 ! your N will be negative ! Mine will be positive =1

Answer 19

so mine is not true? Im still a little iffy on whats going on with these problems. I learned this today...

Answer 20

I meant by yours "ln(1/ep)/ln(10/9)"
we are just dicussing ! so...Take it easy :)

Answer 21

lol? Im easy. Im jsut asking cause im a little lost, and if the way Im doing it is not right im more lost:) Im not heated man. Dont know why it sounds that way but sorry:)

Answer 22

I guess I dont understand why yet, but I just need to think about it more. What about the +1? what is that about?

Answer 23

you know what's this \[\lfloor what's this \rfloor\] ?

Answer 24

nope

Answer 25

ohhh ! This is the reason of your lost !

Answer 26

Im guessing thats only one of the reasons im lost:P

Answer 27

trust me this the only one :) !

Answer 28

so what is it?

Answer 29

wait I looking for a link !

Answer 30

ahh tyvm

Answer 31

integer part means something to you ?!

Answer 32

yeah

Answer 33

that is the integer part ! :)

Answer 34

now ! you agree that we have to choose N as a positive integer ! ?

Answer 35

o that is notion for integer part, sweet. Yes tyvm

Answer 36

notation*

Answer 37

thats handy:)

Answer 38

now ! did you understand !? :) !

Answer 39

yes, forcing N to be positive.

Answer 40

yeah !

Answer 41

I love this stuff:) thankyou for your time.

Answer 42

you're welcome ! just practice more ! you'll be fine !