Easy sequence convergence problem

[jimmypoopins]

Homework Statement

Show that the sequence defined by y_1=1 and y_n+1 = 4-(1/y_n) converges and find the limit.

Homework Equations

thm: a sequence that is monotone and bounded is convergent.

The Attempt at a Solution

i used induction to show that the sequence is increasing and bounded above by 4, so i showed that it is convergent.

i'm having issues showing mathematically that the limit is 4. maybe I'm forgetting something from calc 2?

the book suggests to "take the limit of each side of the recursive equation to explicitly compute lim y_n," but i don't understand how to do that.

any help in the right direction would be gratefully appreciated :)

 

[sutupidmath]

well you first have to show that this sequence is monotonic and bounded, after that show that the lim exists and find it.

y_1=1, y_2=3, y_3=11/3 now use induction to show that this sequence is monotono increasing.

After that try to show that the sequence is upper bounded by 4. Again you have to use induction to prove this, maybe there is any other what but this is what crosess my mind right now.

After that let lim(n-->infinity)(y_n)=L and see what happenes.

now you want to go like this

lim(n->inf)(y_n+1)=4-1/lim(n->inf)(y_n) , then replace it by L, you will get a quadratic eq, solve it for L, and see which result to take?

do u get it?

 

[jimmypoopins]

yeah, i understand. i did the first part and proved that it is monotone and bounded.

i'm confused on the part "lim(n->inf)(y_n+1)=4-1/lim(n->inf)(y_n) , then replace it by L, you will get a quadratic eq, solve it for L, and see which result to take?"

if lim(n->inf)(y_n)=L, then what does lim(n->inf)(y_n+1) equal?

if i substitute in L, i get

lim(n->inf)(y_n+1)=4-1/L. how does that help me?

 

[sutupidmath]

lim(y_n)(n->inf)=L. this means that:
for every epsylon e>0, there exists some N(e)>0 such that for any n>N Iy_n-LI<e
now n+1 is obviously greater than n, right?
what do u think lim(n->inf)(y_n+1) equals now?

 

[jimmypoopins]

ah, since it's an infinite series, then |y_n+1-L|<e also. so i get

L=4-(1/L) and i solve for L.

thanks :)

edit: no, I'm clearly wrong because that implies

L+1/L=4
(L^2+1)/(L+1)=4 isn't satisfied if L=4...

 

[sutupidmath]

yep, and you will end up with two values for L, but since the sequence is monotono increasing you have to chose only one of them, whichever is closest to 4.

 

[jimmypoopins]

L+1/L=4
(L^2+1)/(L+1)=4

is only satisfied if L=0 or L=1, so you're saying the limit is 1? i thought it was 4.

edit: no I'm stupid. sorry. I'm bad at algebra. i got it now, thanks :)

 

[jimmypoopins]

one last question: i get that the equation is satisfied when L=3.7320508... (via calculator)

how can i algebraically solve L=4-(1/L) for L to get an exact answer? i don't think we are allowed to use calculators on homework.

 

[sutupidmath]

one last question: i get that the equation is satisfied when L=3.7320508... (via calculator)

how can i algebraically solve L=4-(1/L) for L to get an exact answer? i don't think we are allowed to use calculators on homework.

well you multiply through by L, and you will get

L^2-4L+1=0, then by applying viettis formulas we get

L_1,2=[-b^2+-sqrt(b^2-4ac)]/2a, where a=1, b=-4, c=1
and your final answer will be 2+sqrt(3)

 

[sutupidmath]

This is just some height school algebra problem, i mean this last part!

 

[sutupidmath]

edit: no, I'm clearly wrong because that implies

L+1/L=4
(L^2+1)/(L+1)=4 isn't satisfied if L=4...

who said that the limit will be 4 at first place?