Question

the fibbonacci sequence is defined by an+1 = an+1 +an for n more than and equal to 1 where a1= 1, a2=2

find limit an+1/an assuming it exists

Answer 1

oh i see, it is
\[\lim_{n\to \infty}\frac{a_{n+1}}{a_n}\] a famous number

Answer 2

yup but i did not understand this chapter much in the class..can you help me?

Answer 3

sure i think

Answer 4

rewrite the numerator \(a_n+a_{n-1}\)

Answer 5

\[\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{a_n+a_{n-1}}{a_n}\]

Answer 6

first off is that clear?

Answer 7

sorry..i forgot to give the hint..an+1/an = an+2/an+1

Answer 8

no need

Answer 9

is it clear what i wrote?

Answer 10

that is just step one
we do a small bit of algebra, then solve a quadratic equation
it will not take long

Answer 11

where did u get an + an - 1?

Answer 12

that is how the fibonacci sequence is defined
each term is the sum of the two preceding terms
so for example
\[a_8=a_7+a_6\] and \[a_{101}=a_{100}+a_{99}\] and
\[a_{n+1}=a_n+a_{n-1}\]

Answer 13

clear or no?

Answer 14

clear :)

Answer 15

ok so now we are at this step
\[\lim_{n\to \infty}\frac{a_n+a_{n-1}}{a_n}\] we we break in to two parts
\[\lim_{n\to \infty}1+\frac{a_{n-1}}{a_n}\]

Answer 16

let me now if that is ok, it is algebra

Answer 17

ok i got it..next?

Answer 18

ok now comes a small thinking part

Answer 19

\[\frac{a_{n+1}}{a_n}\] is the next term over the previous one, wheras \[\frac{a_{n-1}}{a_n}\] is the previous term over the next term

Answer 20

clear?

Answer 21

ok clear

Answer 22

so if
\[\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=x\] (here we assume it exists) then
\[\lim_{n\to \infty}\frac{a_{n-1}}{a_n}=\frac{1}{x}\]

Answer 23

as one is the reciprocal of the other, at least in the limit

Answer 24

only one step to go
so far so good?

Answer 25

no..i don't understand at 1/ x

Answer 26

ok that is really the only part that requires thought, so i have to say it in english

Answer 27

\[\frac{a_{n}}{a_{n+1}}\] is the reciprocal of \[\frac{a_{n+1}}{a_n}\] should be clear

Answer 28

now as you take the limit as n goes to infinity there is no difference between
\[\lim_{n\to \infty}\frac{a_n}{a_{n+1}}\] and
\[\lim_{n\to \infty}\frac{a_{n-1}}{a_n}\]

Answer 29

what is says in english is the limit of the previous term over the next term

Answer 30

the subscripts are unimportant except that the indicate that you are taking the limit of the ration of one term over the previous one
i hope that is clear

Answer 31

ohhh i see..there is no difference as we use infinity right?

Answer 32

right

Answer 33

you replace \(n\) by \(n+1\) everywhere and nothing changes

Answer 34

ok so now what do we have? putting it together we have
\[\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=1+\lim_{n\to \infty}\frac{a_{n-1}}{a_n}\] and assuming the limit exists and is say \(x\) this equation becomes
\[x=1+\frac{1}{x}\]

Answer 35

then do quadratic right?

Answer 36

and now i have to run, but this is a simple enough equation to solve
you get
\[x^2=x+1\] or
\[x^2-x-1=0\] which will solve using the quadratic formula, get a famous number called "the golden ration" sometimes denoted as \(\phi\)

Answer 37

yeah, a quadratic
actually you get two answers, pick the one that is greater than one, since the limit obviously is
google "golden ratio" and you will get more information than you can use

Answer 38

ok i got it..thank you very very much for helping..

Answer 39

yw