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Homework Help: When does lim a_n x^n exist?

  1. Aug 14, 2010 #1
    The problem statement, all variables and given/known data
    Let an be a sequence of real numbers. For what values of x does lim anxn exist?

    The attempt at a solution
    Let us suppose that lim anxn exist and is equal to b. What can we say about x? Hmm...there is a monotonic subsequence that converges to b, say [itex]a_{k_n}x^{k_n}[/itex]. If this is an increasing sequence, we have that

    a_{k_n}x^{k_n} \le a_{k_{n+1}}x^{k_{n+1}}

    or equivalently

    \frac{a_{k_n}}{a_{k_{n+1}}} \le x^{k_{n+1} - k_n}

    Unfortunately I don't get an inequality in terms of x alone. How do I proceed from here? Perhaps I need a further assumption, like [itex]k_{n+1} - k_n = 1[/itex] for all n?
  2. jcsd
  3. Aug 14, 2010 #2


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    its going to depend on your a_n... if a_n =0 for all n, it exists for every x
  4. Aug 14, 2010 #3


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    sorry, actually I misread sequence

    if the limit exists, and is b say, then for any e>0 you can choose N such that for all n>N
    [tex] |a_nx^n -b|<e [/tex]

    you could then think about the beahviour of x when x>|1| and otherwise

    though as mentioned previously it will depend on teh sequnce [itex] a_n [/itex] , is it for any [itex] a_n [/itex] or are there any contraints on the [itex] a_n [/itex] ?
  5. Aug 14, 2010 #4
    There are no constraints on an. I do know of a couple that might shed some light on this general situation. As you previously mentioned, if the an are all 0 (or are eventually all 0), then the limit exists for all n. If the an are eventually all some nonzero constant, then we know the limit exists for all |x| < 1. Now I believe that if c = lim |an/an+1|, then the limit exists for all |x| < c. This is something I have yet to prove though.
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