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Homework Help: Putnam problem from 1949 - lim sup

  1. Aug 13, 2011 #1

    jbunniii

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    This is a nice Putnam problem from the 1949 exam. It also appears as challenge problem 2.14.16 in Thomson, Bruckner and Bruckner, Elementary Real Analysis. I think I solved it correctly, but it seemed a little too easy so if anyone would like to check my solution, I would appreciate it.

    1. The problem statement, all variables and given/known data

    Let [itex](a_n)[/itex] be an arbitrary sequence of real, positive numbers. Show that

    [tex]\limsup_{n \rightarrow \infty} \left(\frac{a_1 + a_{n+1}}{a_n}\right)^n \geq e[/tex]

    2. Relevant equations

    [tex]e = \lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^n[/tex]

    3. The attempt at a solution

    Suppose not. Then there exists [itex]N \in \mathbb{N}[/itex] and [itex]\alpha \in \mathbb{R}[/itex] such that

    [tex]\left(\frac{a_1 + a_{n+1}}{a_n}\right)^n \leq \alpha < e[/tex]

    for all [itex]n \geq N[/itex]. Also, there exists [itex]M \in \mathbb{N}[/itex] such that

    [tex]\left(1 + \frac{1}{n}\right)^n > \alpha[/tex]

    for all [itex]n \geq M[/itex]. Therefore, for all [itex]n \geq \max(N,M)[/itex], we have

    [tex]0 < \left(\frac{a_1 + a_{n+1}}{a_n}\right)^n < \left(1 + \frac{1}{n}\right)^n[/tex]

    and therefore

    [tex]0 < \frac{a_1 + a_{n+1}}{a_n} < 1 + \frac{1}{n} = \frac{n+1}{n}[/tex]

    After rearrangement, this is equivalent to

    [tex]\frac{a_n}{n} - \frac{a_{n+1}}{n+1} > \frac{a_1}{n+1}[/tex]

    This inequality holds for all [itex]n \geq \max(M,N)[/itex]. If I write out the inequality for [itex]n[/itex] through [itex]n+k-1[/itex], and then sum these inequalities, then the LHS telescopes, and I obtain

    [tex]\frac{a_n}{n} - \frac{a_{n+k}}{n+k} > \sum_{j = n}^{n+k-1}\frac{a_1}{j+1}[/tex]

    This holds for any positive integer [itex]k[/itex]. If I hold [itex]n[/itex] fixed and let [itex]k[/itex] grow large, the right hand side is unbounded. In particular, there is some [itex]k[/itex] so large that the right-hand side is larger than [itex]a_n/n[/itex]. Thus for that [itex]k[/itex], we have

    [tex]\frac{a_n}{n} - \frac{a_{n+k}}{n+k} > \frac{a_n}{n}[/tex]

    or equivalently

    [tex]\frac{a_{n+k}}{n+k} < 0[/tex]

    and thus

    [tex]a_{n+k} < 0[/tex]

    This is a contradiction because the sequence contains only positive numbers.
     
    Last edited: Aug 13, 2011
  2. jcsd
  3. Aug 13, 2011 #2
    It looks right to me.
     
  4. Aug 13, 2011 #3
    I'm happy with it.
     
  5. Aug 13, 2011 #4

    jbunniii

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    Cool, thanks for taking the time to check it out.
     
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