Question

calc-geometric sequence

Answer 1

Find the values of x for which the series converges.
Find the sum of the series for those values of x.
\[\sum_{n=0}^{\infty} \frac{ (x+6)^n }{ 2^n}\]

Answer 2

ratio test for this one

Answer 3

first write \[|\frac{a_{n+1}}{a_n}|\] then do a raft of cancellation
let me know what you get

Answer 4

or if it is not clear, let me know

Answer 5

Why not root test?

Answer 6

because it is a set up for the ratio text
every damn thing will cancel

Answer 7

im not sure how to do it..

Answer 8

\[a_n=\frac{(x-6)^n}{2^n}\]

Answer 9

replace \(n\) by \(n+1\) get \[a_{n+1}=\frac{(x+6)^{n+1}}{2^{n+1}}\]

Answer 10

i made a typo, but you get the idea right?

Answer 11

so \[|\frac{a_{n+1}}{a_n}|=|\frac{(x+6)^{n+1}}{2^{n+1}}\times \frac{2^n}{(x+6)^n}|\]

Answer 12

now before worrying about the limit as \(n\to \infty\) do a bunch of cancellation

Answer 13

let me know if any step is not clear

Answer 14

\[= \frac{ x+6 }{ 2}\]? :/

Answer 15

oui

Answer 16

well actually \[\frac{1}{2}|x
+6|\]

Answer 17

you need the absolute value around the \(x+6\) part

Answer 18

now clearly \[\lim_{n\to \infty}\frac{1}{2}|x+6|=\frac{1}{2}|x+6|\] because there is no \(n\) left

Answer 19

do you know what comes next ?

Answer 20

no

Answer 21

ok ratio test says series will converge if \[\lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|<1\]

Answer 22

we found the limit to be \[\frac{1}{2}|x+6|\] so set \[\frac{1}{2}|x+6|<1\] solve for \(x\)

Answer 23

that part should be pretty easy
let me know if you get stuck
still not done however

Answer 24

x<-4 ?

Answer 25

one side

Answer 26

\[\frac{1}{2}|x+6|<1\\
|x+6|<2\\
-2<x+6<2\]

Answer 27

\[-8<x<-4\]

Answer 28

btw \(|x+6|<2\) says "radius of convergence is 2"

Answer 29

ready for the next part?

Answer 30

yes

Answer 31

we have to test the endpoints of the interval, because the ratio test does not say what happens if the limit is 1
so first lets put \(x=-8\) and see if this sucker converges

Answer 32

this is not nearly as hard as it may look
if we replace \(x\) by \(-8\) in \(x+6\) we get \(-8+6=-2\)
so the sum would be \[\sum\frac{(-2)^n}{2^n}\]

Answer 33

but \[\frac{-2)^n}{2^n}=\left(\frac{-2}{2}\right)^n=(-1)^n\]
so you would have \[\sum(-1)^n\] converge or no?

Answer 34

i dont think so

Answer 35

of course not
the terms to not even to to zero \[1-1+1-+1-1+...\]

Answer 36

now try it with \(x=-4\)

Answer 37

2^n/2^n

Answer 38

yeah until you reduce and get 1

Answer 39

ok

Answer 40

so no point in asking if \[\sum 1\] converges right?

Answer 41

it does

Answer 42

???

Answer 43

\[1+1+1+1+...\]

Answer 44

oh..it doesnt :/

Answer 45

course not
so the interval is open \[-8<x<-4\]

Answer 46

gotcha

Answer 47

finito (i hope)
this was a long one, but all the steps are there, don't think we skipped any

Answer 48

im sorry, theres a second part to the question, to find the sum

Answer 49

@terenzreignz was right actually root test might have worked nicely here

Answer 50

ok find the sum damn

Answer 51

so lets put say \(x=-6\) and see what we get

Answer 52

ok that is a stupid choice, get zero nvm
lets try \(x=-7\)

Answer 53

you would have \[\sum\frac{(-1)^n}{2^n}\] i.e. \[\sum\left(-\frac{1}{2}\right)^n\] a geometric sequence

Answer 54

sum is \[\frac{1}{1-(-\frac{1}{2})}\]

Answer 55

repeat with \(x+6\) instead of \(-7+6\)

Answer 56

im sorry, what do you mean by x+6 instead of -7+6?

Answer 57

\[\frac{ x+6 }{ 1-\frac{ x+6 }{ 2 } }\]

Answer 58

like that?

Answer 59

yes then some algebra to make it look less silly

Answer 60

i.e multiply top and bottom by 2 maybe

Answer 61

\[\frac{ 2(x+6) }{ 2-\left| x+6 \right| }\]

Answer 62

wait a sec maybe i messed up when i said "yes"

Answer 63

yeah i messed up
numerator should also by \(\frac{x+6}{2}\)

Answer 64

\[\frac{\frac{x+6}{2}}{1-\frac{x+6}{2}}\]

Answer 65

or \[\frac{x+6}{2-(x+6)}\] which you can also clean up a bit

Answer 66

whew now are we done?

Answer 67

is that the sum?

Answer 68

what do i do with the 1/ (1+(1/2)) ?

Answer 69

yes that is the sum
i just picked \(x=-7\) as an example

Answer 70

so you would see what you get when you use \(x\)

Answer 71

we can check you get the same number if you like

Answer 72

hmm it says that answer is incorrect

Answer 73

really ??

Answer 74

did you put in \[\frac{x+6}{-x-4}\]?

Answer 75

i put (x+6) / 2-(x+6)

Answer 76

maybe i made a mistake

Answer 77

got logged out, sorry

Answer 78

ooh k i see the problem

Answer 79

\[\sum r^n=\frac{1}{1-r}\]so \[\sum\left(\frac{x+6}{2}\right)^n=\frac{1}{1-\frac{x+6}{2}}\]

Answer 80

multiply top and bottom by 2, get \[\frac{2}{2-(x+6)}=\frac{2}{-x-4}\]

Answer 81

ok thats right. thank you:)

Answer 82

whew !