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Determining the convergence of a recursive sequence

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Given the sequence

    [tex] L_n = Q \cdot (1 + \frac{r}{n}} )^n [/tex]
    where [tex] Q > 0, r > 0 [/tex]

    show that it either converges or diverges.

    2. Relevant equations

    Restating the sequence as a recurrence could (maybe) be of help:

    [tex] L_{n+1} = L_n \cdot (1 + \frac{r}{n}} ) [/tex] and [tex] L_1 = Q \cdot (1 + \frac{r}{1}}) [/tex]

    3. The attempt at a solution

    I've tried two different approaches:
    1. Using the epsilon-definition of convergence:


      Using Maple I've found the limit of the sequence to be [tex] e^r \cdot Q [/tex], so there exists a [tex] N \in \mathbb{R} [/tex] such that

      [tex] \vert Q \cdot (1 + \frac{r}{n}} )^n - e^r \cdot Q \vert < \epsilon [/tex]
      for all [tex] n \geq N [/tex].

      The problem with this method is, that I can't figure out what to do upon rewriting the above inequality:
      [tex] \vert \frac{(n+r)^n}{n^n} - e^r \vert < \frac{\epsilon}{Q} [/tex]

      Ideally, I should either solve for [tex]n[/tex] , thus allowing me to find a suitable [tex]\epsilon[/tex] or reduce the inequality to some form where choosing and [tex]\epsilon[/tex] would be obvious. But it seems impossible due to the exponentiation.


    2. Using that a sequence converges iff it is monotonous and has a supremum/infimum
      Therefore I've thought of restating the sequence as a recurrence and showing that it is monotunous and has a supremum.
      Showing that the sequence is monotonous is easy:

      [tex]r > 0[/tex], so [tex] (1 + \frac{r}{n}}) > 1 [/tex], so [tex] L_{n+1} > L_n [/tex].

      The second part seems unsolvable, though.
      I've tried to use induction to show that the sequence has a supremum.

      Base step:
      I guess that proving [tex]Q \cdot (1 + r) < Q \cdot e^r [/tex] is easy, and so I haven't yet spent time on proving it.

      What I have spent time on, however, is the inductive step:
      Assuming that [tex]L_k < Q \cdot e^r[/tex], I need to prove that [tex]L_{k+1} < Q \cdot e^r[/tex].

      This problem reduces to showing the following:
      [tex](1 + \frac{r}{k})^k < e^r [/tex] iff [tex](1 + \frac{r}{k})^{k+1} < e^r[/tex]

      I have no idea how one would go about proving that the left side of the 'iff' implies the right side of the 'iff'

    I would greatly appreciate if anyone could help me making some advances using either method 1 or method 2 - or an entirely different method.
     
    Last edited: Sep 13, 2010
  2. jcsd
  3. Sep 13, 2010 #2
    It's a standard limit, Q exp(r).
     
  4. Sep 13, 2010 #3
    Thanks for you quick answer. Could you elaborate on it being a "standard limit"? -I don't see how it will help me in showing whether or not the sequence converges..
     
  5. Sep 13, 2010 #4
    That standard limit is [tex]\lim_{n\rightarrow\infty}\left(1 + \frac{x}{n}\right)^n = e^x[/tex]

    If you wanted to go this route and use the limit, I think you could use the squeeze theorem to prove the limit.
     
  6. Sep 14, 2010 #5
    I doubt that to be the way which I am to solve that homework problem given that this standard limit is not mentioned in my textbook nor have we yet covered the squeeze theorem.
     
  7. Sep 14, 2010 #6
    We really just need to show (1 + 1/n) ^ n converges as Q is a constant and the r can be written as an exponent.

    You know the sequence is increasing you just need to show it bounded by some M. It turns out 3 works. Use the binomial theorem to show each Ln is less then 3. This is probably what your prof wants but it only says Ln has some least upper bound (which is e).


    alternatively if you can use the properties of logs and know log'(1) = 1 then I'd use this derivation from Spivak which gives e ^ x, show:

    1) lim y->0 log(1+y)/y

    2)lim x -> inf x*log(1 + (1/x))

    3)e = lim x->inf [1+ (1/x)]^x (bring the limit outside)

    then use a substitution to show lim n -> inf [1+ (r/x)]^x = e^r. The Q is just a constant.
     
    Last edited: Sep 14, 2010
  8. Sep 14, 2010 #7
    Thanks for your elaborate answer, deluks917.

    With regards to your alternative way to solve the problem, I am wondering whether it is logically sound.. Since I am to simply determine whether the sequence converges or diverges it seems incorrect to set out on determining the limit, thereby assuming that it converges. Should you not first conclude that it converges?

    With regards to using the binomial theorem I have, upon reading about it, found it to be a perfect way to proof that it is bounded. However, we have not covered it in class and so I am wondering whether I can use it. Is there no other way to show that it is bounded than using the binomial theorem?
     
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