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Math Help - Having some trouble with a limit

  1. #1
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    Having some trouble with a limit

    Dear all,

    I'm having some trouble with the following limit:
    n(x^n) as n tends to infinity, knowing that |x| < 1
    I need it to calculate a series.

    Please note that I am not allowed to use exp and ln as we haven't covered that in the syllabus yet. However, I'd be glad to know of a solution using these functions too.

    Thanks in advance.
    Yack

    PS: this is my first post here, sorry if I left this message in the wrong place.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Yack View Post
    Dear all,

    I'm having some trouble with the following limit:
    n(x^n) as n tends to infinity, knowing that |x| < 1
    I need it to calculate a series.

    Please note that I am not allowed to use exp and ln as we haven't covered that in the syllabus yet. However, I'd be glad to know of a solution using these functions too.

    Thanks in advance.
    Yack

    PS: this is my first post here, sorry if I left this message in the wrong place.
    what do you want exactly? to calculate a limit, or to tell if a series converges? it looks to me like the root test is your best bet for the latter. the ratio test can work too
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    what do you want exactly? to calculate a limit, or to tell if a series converges? it looks to me like the root test is your best bet for the latter. the ratio test can work too

    I'm trying to calculate the limit of the sequence n*x^n.
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  4. #4
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    Quote Originally Posted by Yack View Post
    I'm trying to calculate the limit of the sequence n*x^n.
    Look at the root, |nx^n|^{1/n} = n^{1/n} |x| now n^{1/n} \to 1 thus n^{1/n} |x| \to |x|. Thus, if |x|<1 then the sequence limit is 0. If |x|>1 then the sequence limit diverges. And if |x|=1 then we have two cases to check x=\pm 1 in either case the limit does not exist.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Look at the root, |nx^n|^{1/n} = n^{1/n} |x| now n^{1/n} \to 1 thus n^{1/n} |x| \to |x|. Thus, if |x|<1 then the sequence limit is 0. If |x|>1 then the sequence limit diverges. And if |x|=1 then we have two cases to check x=\pm 1 in either case the limit does not exist.
    Thanks a lot. This makes it much more obvious. However I'm having trouble to find a formal proof (using Cauchy sequences, the squeeze theorem, or the definition of limits).

    Yack
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  6. #6
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    Quote Originally Posted by Yack View Post
    Thanks a lot. This makes it much more obvious. However I'm having trouble to find a formal proof (using Cauchy sequences, the squeeze theorem, or the definition of limits).

    Yack
    Theorem: Let a_n be a sequence such that |a_n|^{1/n} is a convergent sequence with limit a. If a<1 then the sequence a_n \to 0 and if a>1 then a_n is divergent.

    Proof: Say 0\leq a<1 (it cannot be that a<0 because this is a sequence of non-negative terms) and \lim ~ |a_n|^{1/n} = a then \left| |a_n|^{1/n} - a \right| < \epsilon for n\geq N. This means,  a - \epsilon < |a_n|^{1/n} < a+\epsilon for n\geq N, choose \epsilon so small that a+\epsilon < 1 (which is possible since a<1). Then it means 0\leq |a_n|^{1/n} \leq (a+\epsilon)\implies 0\leq |a_n| \leq (a+\epsilon)^n \mbox{ for }n\geq N by the Squeeze theorem \lim ~ |a_n| = 0 because (a+\epsilon)^n \to 0. Thus, \lim ~ a_n = 0. Q.E.D.

    The case with a>1, is similar, i.e. pick an \epsilon so small that a-\epsilon > 1 and then (a-\epsilon)^n \leq |a_n|.
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