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Sequences, induction help

  1. Apr 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Let a_n be defined recursively by
    [tex]a_{1}[/tex]=1, [tex]a_{n+1}[/tex]=sqrt(6+[tex]a_{n}[/tex]) (n=1,2,3,...).
    Show that lim n->infinity [tex]a_{n}[/tex] exists and find its value

    3. The attempt at a solution
    Observe that [tex]a_{2}[/tex]=[tex]\sqrt{6+1}[/tex]=[tex]\sqrt{7}[/tex] > [tex]a_{1}[/tex]. If [tex]a_{k+1}[/tex] > [tex]a_{k}[/tex], then [tex]a_{k+2}[/tex] = [tex]\sqrt{6+a_{k+1}}[/tex] > [tex]\sqrt{6+a_k}[/tex] = [tex]a_{k+1}[/tex], so {[tex]a_{n}[/tex]} is increasing by induction. I get that part, its the next part I'm a little confused about:

    Now, observe that [tex]a_{1}[/tex]=1 < 3. If [tex]a_{k}[/tex] < 3, then [tex]a_{k+1}[/tex]=[tex]\sqrt{6+a_k}[/tex] < [tex]\sqrt{6+3}[/tex], so [tex]a_{n}[/tex] < 3 for every n by induction. How does this show that for all n, the value will be less than 3? Sorry, I'm not really sure how it shows that if you keep going on in the sequence it won't go past 3 eventually.
     
  2. jcsd
  3. Apr 4, 2010 #2

    tiny-tim

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    Hi Linday12! :wink:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    But you've proved it!

    If an < 3, then an+1 < 3 …

    what is worrying you about that? :smile:
     
  4. Apr 5, 2010 #3
    Sorry, I wasn't clear enough. I'm just not seeing how it shows that. I've memorized it and am not really worried about getting it wrong, but it's really not helpful if I don't understand that it keeps under the limit. I guess my induction skills are lacking.

    Ok, so if ak < 3, then ak+1 = [tex]\sqrt{6+a_{k}}[/tex] < [tex]\sqrt{6+3}[/tex], so an < 3 for every n by induction.

    What I'm not seeing is how this shows that it is below 3 for n. For example, as you keep going up to something like a100, how do you know that it won't add up to enough to push over 3. My (wrongful) reasoning is that if you keep adding more and more, it finally accumulates enough to surpass 3 and go to infinity. I can't seem to see how it approaches the limit. :frown:
     
  5. Apr 5, 2010 #4
    What you have just proved is: [itex](\star)[/itex] the validity of the statement "[itex]a_k<3[/itex]" implies the validity of the statement "[itex]a_{k+1}<3[/itex]." What this allows you to do is essentially re-use [itex](\star)[/itex] over and over. We know that [itex]a_1<3[/itex]. By [itex](\star)[/itex], the validity of [itex]a_1 < 3[/itex] implies the validity of [itex]a_2 < 3[/itex]. But, again by [itex](\star)[/itex], the validity of [itex]a_2<3[/itex] implies the validity of [itex]a_3<3[/itex] and so on.

    In induction proofs, what you're proving is that the validity of the k-th case implies the validity of the (k+1)-th case. Along with the base case, this allows you to conclude the statement is true for all values of k.
     
  6. Apr 6, 2010 #5

    tiny-tim

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    Hi Linday12! :smile:

    (just got up :zzz: …)
    Nothing is accumulating, there's no sum, each an is on its own.
    Yes, you need to convince yourself that induction is valid.

    Go over what rs1n :smile: has said, and try to convince yourself that it makes sense (it does!! :biggrin:).

    Then try to apply it in other examples (such as an greater than 3 in this case). :smile:
     
  7. Apr 11, 2010 #6
    Thank you both very much! Those two posts cleared up my problems with it. I was thinking of it in an accumulating way :uhh: (I have no idea why, series and sequence mix up I guess :blushing:), so now it makes sense.

    :cool: Thanks again!
     
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