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Easy sequence convergence problem

  1. Feb 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that the sequence defined by y_1=1 and y_n+1 = 4-(1/y_n) converges and find the limit.


    2. Relevant equations
    thm: a sequence that is monotone and bounded is convergent.


    3. 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 :)
     
  2. jcsd
  3. Feb 17, 2008 #2
    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?
     
  4. Feb 17, 2008 #3
    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?
     
  5. Feb 17, 2008 #4
    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?
     
  6. Feb 17, 2008 #5
    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...
     
    Last edited: Feb 17, 2008
  7. Feb 17, 2008 #6
    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.
     
  8. Feb 17, 2008 #7
    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 :)
     
    Last edited: Feb 17, 2008
  9. Feb 17, 2008 #8
    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.
     
  10. Feb 17, 2008 #9
    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)
     
  11. Feb 17, 2008 #10
    This is just some hight school algebra problem, i mean this last part!
     
  12. Feb 17, 2008 #11
    who said that the limit will be 4 at first place????????
     
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