Find limits of a ratio of sequences...?

Question

(1+sqrt3)^(2n+1)=an+bn.squareroot3, an, bn are sequences,the dot represents the multiplication operation there, and an, bn belong to Q, so we must find lim (an/bn) when n goes to infinity. Sorry but yahoo! answers cannot allow me to write as easy as i do at the blackboard or on my notebok. :) looing forward for answers. If something is unclear pls post question and i will try to explain better.

This is not homework. I am working math problems on my own for this year's exam. If answer's like julia's are received i will report abuse. I am not asking this just to waste my time...

Yes, an and bn are "a subscript n" and "b subscript n". squarer is square root but yahoo answers did not show it (reason don't know why) :D

(1+sqrt3)^(2n+1) means: one plus squareroot three, everything to the power two times n plus one :D

i can upload a photo of the exercise but the problem is where...i don't know any site and i hope yahoo answers does not forbid links :) if u can help me lucy.

ok answer found no need to answer this...

by using newton binom u can prove that:
(1-sqrt3)^(2n+1)=an-bn x sqrt3
If u add the two relations u have u will obtain an because some expressions simplify...in the same manner by replacing an u can find bn. so by using lim q^n =0 when n goes to infinity and q belongs to (-1,1)...because 1-sqrt3 or sqrt3-1 belong to that interval....u can obtain that the limit is equal to sqrt3....ANSWER FOUND

Answer 1

a_0 = b_0 = 1.

(a_n+1) + (b_n+1)√3) = (a_n + b_n√3)(1 + √3)² = (a_n + b_n√3)(4 + 2√3) =
(4a_n + 6b_n) + (2a_n + 4b_n)√3

So a_n+1/b_n+1 = (4a_n + 6b_n)/(2a_n + 4b_n) = (2a_n + 3b_n)/(a_n + 2b_n)

(2a_n + 3b_n)/(a_n + 2b_n) = (2a_n + 4b_n - b_n)/(a_n + 2b_n)

(2a_n + 4b_n - b_n)/(a_n + 2b_n) = (2(a_n + 2b_n) - b_n))/(a_n + 2b_n) = 2 - b_n/(a_n + 2b_n)

That says the sequence (a_n/b_n) is bounded above by 2.
Since b_n and a_n are > 0, b_n/(a_n + 2b_n) < 1/2 so the ration is bounded below by 1.5

I ran some calculations using Excel and it looks like the limit is √3 but I can't see how to get there from here.

Addenda:
Let c_n = a_n/b_n. So far we have

c_n+1 = 2 - b_n/(a_n + 2b_n)
Divide the right hand side top and bottom by b_n and you get

c_n+1 = 2 - 1/(a_n/b_n + 2) = 2 - 1/(c_n + 2)

Note that if you set c_n = √3, then c_n+1 = √3

Answer 2

(a(n)+sqrt 3 (b(n))=(a(n-1)+sqrt3 b(n-1))(1+sqrt3)
you get a(n)=a(n-1)+3b(n-1)
and b(n)=a(n-1)+ b(n-1)

but b(n-1) = (a(n)-a(n-1))/3 so
b(n)= a(n-1)+(a(n)-a(n-1))/3 = 2a(n-1)/3 +
a(n)/3
so lim (b(n)/a(n))=lim(2a(n-1)/3 a(n)) + 1/3
so your problem is equivalent to find
lim a(n-1)/a(n)
i am stuck
OR.... (1-sqrt 3 )^(2n+1) = a(n)- b(n)sqrt 3
if you add with the first equation
(1-sqrt 3 )^(2n+1) + (1+sqrt 3 )^(2n+1) = 2 a(n)
and
(1+sqrt 3 )^(2n+1) - (1-sqrt 3 )^(2n+1) = 2 b(n) sqrt 3
so you need to find
((1-sqrt 3 )^(2n+1) + (1+sqrt 3 )^(2n+1) )/(((1+sqrt 3 )^(2n+1) - (1-sqrt 3 )^(2n+1) )sqrt 3) when n goes to infinity
by a standard procedure you obtain limit
1/sqrt 3

Answer 3

(1+sqrt3)^(2n+1)= 1 + (2n+1).sqrt3+((2n+1)(2n)).3/2+...... So what you mean by an and bn?
Try to write your problem in more details

Answer 4

it 'd be easier to answer if i could see clearly, i have some problems with that type of print i can't think very well unless its on paper, here's an idea if you have a scanner scan the problem well, im sorry im wasting ur time good luck with ur exam

Answer 5

Ummm. I think you should do your own homework and ask for help from a tutor or your instructor if you need it. :)