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Prove that this sequence is bounded

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data

    A sequence (an: [tex]n \in N[/tex]) is defined by an= (2n+3)/(3n+6) for all [tex]n \in N[/tex].

    (a) Prove that this sequence is bounded above by 2/3;

    (b) Prove that the sequence (an: [tex]n \in N[/tex]) is monotonely increasing by showing that 0<an+1-an for all [tex]n \in N[/tex].

    3. The attempt at a solution

    I need to write a perfect proof for this question, I made an attempt but I'm not sure if I'm using the right method and whether my proof is valid or makes any sense...

    My attempt for part (a):

    I want to use the method of induction. For this I need to show that (A) [tex]1 \in S[/tex], then (B) if a number [tex]n \in S[/tex] then so is the number after n.

    If k is some natural number in the sequence, for k=1:

    [tex]a_{n}=\frac{2.1+3}{3.1+6} = \frac{5}{9} \leq \frac{2}{3}[/tex]

    Now, we have shown that [tex]a_{k} \leq \frac{2}{3}[/tex] and also premise A

    ak+1 = (2.2+3)/(3.2+6)= 7/12 ≤ 2/3

    Therefore an ≤ 2/3 for all [tex]n \in N[/tex].

    I appreciate some guidance. Thank you.
     
  2. jcsd
  3. Apr 5, 2009 #2
    Re: Proofs

    Have you tried manipulating the expression (2n+3)/(3n+6)? Separating the expression into two fractions might help.
     
  4. Apr 5, 2009 #3

    lanedance

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    Re: Proofs

    hi roam

    i don't think you have shown that by induction, you have shown it is true for n=1 and n = 2

    induction requires showing its true for n=1 and then assume its true for n, use that condition to show it is true for n+1, thus inducting it is true for all n

    looking at ways other than induction, it could be enough to show b) first and then look at the limit as n goes to infinty...

    otherwise for arbitrary n, you could assume a(n)>2/3 and check the condition required on n for this to be true, looking for a contradiction
     
  5. Apr 5, 2009 #4
    Re: Proofs

    Hi!

    That's true but if I want to use induction, I'm not sure how to do the last step:

    Our proposition P(n) is: an ≤ 2/3 for all [tex]n \in N[/tex]

    for n = 1

    [tex]a_{n}=\frac{2.1+3}{3.1+6} = \frac{5}{9} \leq \frac{2}{3}[/tex]

    So, P(n) is true for n=1

    Suppose that P(k) is true, we shall try to show that P(k+1) is true:

    [tex]a_{k+1}=\frac{2(k+1)+3}{3(k+1)+6}[/tex]

    I need some help here because i'm not sure what to do after this...
     
  6. Apr 5, 2009 #5

    lanedance

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    Re: Proofs

    i think the point is you don't really need induction here, following on from cordyceps, you can just re-arrange the fraction

    [tex] a_n = \frac{2.n+3}{3.n+6} = \frac{2.n}{3.n+6} + \frac{3}{3.n+6}[/tex]
    [tex] = \frac{2}{3} (\frac{n}{n+2}) + \frac{1}{n+2}[/tex]
    [tex] = \frac{2}{3} (\frac{n}{n+2} + \frac{3/2}{n+2})[/tex]
    [tex] = \frac{2}{3} (\frac{n + 3/2}{n+2}) [/tex]

    which is <2/3 for all n>0
     
  7. Apr 6, 2009 #6
    Re: Proofs

    So, separating out the expression as you did will be sufficient as a proof for part (a)? (To me it makes sense)


    Btw, does my first post answer part (b)?

    In that post I showed it's true for n=1 and n = 2 but not for all [tex]n \in N[/tex] and that's the problem. Do I need the induction here to show that 0<an+1-an (and hence the series is monotonely increasing)?
     
  8. Apr 6, 2009 #7

    lanedance

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    Re: Proofs

    i don't think you will need to induct... i would first try writing the expression out and see if it is obvious to show it is true for all n
     
  9. Apr 6, 2009 #8

    CompuChip

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    Re: Proofs

    Induction is indeed a very complicated way, because it is not easy to get the right estimate.
    Personally I like the approach where you first solve b - which involves some algebra but is otherwise rather straightforward - and then show a using b and the limit of the terms as n goes to infinity.
     
  10. Apr 6, 2009 #9
    Re: Proofs

    I'm a little confused right now, I mean we already know that [tex]\frac{2}{3} (\frac{n + 3/2}{n+2}) < \frac{2}{3}[/tex] & that sort of answers part (a), doesn't it? For part (b) I have to show that an+1>an, I've already tested the series for n=1 and n=2 but not for all n.

    The limit (let's call it L) of the terms as n->∞ is 2/3 but I don't know how that helps in this case. :rolleyes:
     
  11. Apr 6, 2009 #10

    lanedance

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    Re: Proofs

    Compuchip is suggesting a method for the whole problem that was mentioned earlier, if you show b) the function is monotonically increasing and has limit 2/3, then it can never exceed 2/3, thus proving a)

    However as we have shown a) and want to show b) just write
    [tex] a_{n+1} - a_n [/tex]
    this should simplify to an expression that is always positive, without too much work
     
  12. Apr 6, 2009 #11
    Re: Proofs

    before we find the limit we need to show that it is monotonely increasing, I didn't couldn't show that. I only showed that n=2 > n=1 but didn't show that an+1>an for all n!
     
  13. Apr 6, 2009 #12

    lanedance

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    Re: Proofs

    re-read my last post

    just write
    [tex] a_{n+1} - a_n [/tex]

    it should simply be a matterof re-arranging to show it is positive for all n>0, thus showing the sequence is monotonically increasing

    try it...
     
    Last edited: Apr 7, 2009
  14. Apr 14, 2009 #13
    Re: Proofs

    Yes, I know how to prove part (b) by showing that an+1 - an is some positive expression.

    The problem is here: we must show part a) first and then show part b) (i.e. show that it is bounded above by 2/3 before you show that it is strictly increasing).

    That strategy doesn't work, I'm supposed to answer the questions in order, we can't do b) first and then use the results obtained in b) to prove a).
     
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