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Thread: A limit

  1. #1
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    A limit

    Need serious help on a proof:
    Show that if lim(an/n) = L where L>0, then lim an (n approaches inf) = inf.

    Thanks guys.
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  2. #2
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    Quote Originally Posted by JimmyT View Post
    Need serious help on a proof:
    Show that if lim(an/n) = L where L>0, then lim an (n approaches inf) = inf.

    Thanks guys.
    Informally this is because if:

    $\displaystyle
    \lim_{n \to \infty}a_n/n = L\ >0\ \ \ \dots\ (1)
    $

    for large $\displaystyle n,\ a_n \sim nL$, and as the RHS goes to infinity $\displaystyle a_n$ does as well.

    Now all you have to do is write out what $\displaystyle (1)$ means in full and rearrange it to show that $\displaystyle a_n$
    differers form $\displaystyle nL$ by arbitrarily small amounts for sufficiently large $\displaystyle n$.
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  3. #3
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    I don't understand what you mean when you say "rearrange" it by "arbitrarily small amounts."
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  4. #4
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    $\displaystyle
    \begin{array}{l}
    L > 0\quad \Rightarrow \quad \left( {\exists N} \right)\left[ {n \ge N \Rightarrow \left| {\frac{{a_n }}{n} - L} \right| < \frac{L}{2}} \right] \\
    \Rightarrow \quad \frac{L}{2} < \frac{{a_n }}{n} \\
    \Rightarrow \quad n\frac{L}{2} < a_n \\
    \end{array}
    $
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  5. #5
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    Quote Originally Posted by JimmyT View Post
    I don't understand what you mean when you say "rearrange" it by "arbitrarily small amounts."
    Of course you are right, I don't mean what I say at all do I .

    What I do mean is that for every epsilon>0, there exists an N such that
    for all n>N:

    -epsilon< (a_n)/n-L <epsilon

    so:

    nL-n epsilon< a_n < nL + n epsilon,

    or:

    n(L-epsilon) < a_n < n(L+epsilon)

    so if we choose a small enough epsilon, as n-> infnty a_n is trapped between
    two sequences both of which go to infinity.

    RonL
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