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Math Help - Tricky limit

  1. #1
    Junior Member raheel88's Avatar
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    Tricky limit

    Hi all,

    I'm trying toevaluate the following limit;

    limit as n --> oo (n-1)*(1/2)^(n-1)

    My intuition tells me that it should be zero because (1/2)^(n-1) approaches zero faster than (n-1) approaches infinity...but I'm not sure how to show this rigorously.

    Can anyone help?
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  2. #2
    Senior Member
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    Hi.

    Quote Originally Posted by raheel88 View Post
    Hi all,

    I'm trying toevaluate the following limit;

    limit as n --> oo (n-1)*(1/2)^(n-1)

    My intuition tells me that it should be zero because (1/2)^(n-1) approaches zero faster than (n-1) approaches infinity...but I'm not sure how to show this rigorously.

    Can anyone help?
    This is correct.

    Do you know that a^x = e^{LN(a)*x} ?

    I think that's all you need to know, because we have

    (n-1)*(1/2)^{(n-1)} = (n-1)*e^{\displaystyle LN(0.5)*(n-1)} = (n-1)e^{\displaystyle -0.69*(n-1)}

    And we are done with the same argument as above. exp(-x) tends faster to zero than x to infty. (Do you know this result? Are you allowed to use it?)

    Rapha
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  3. #3
    Junior Member raheel88's Avatar
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    Quote Originally Posted by Rapha View Post
    Hi.



    This is correct.

    Do you know that a^x = e^{LN(a)*x} ?

    I think that's all you need to know, because we have

    (n-1)*(1/2)^{(n-1)} = (n-1)*e^{\displaystyle LN(0.5)*(n-1)} = (n-1)e^{\displaystyle -0.69*(n-1)}

    And we are done with the same argument as above. exp(-x) tends faster to zero than x to infty. (Do you know this result? Are you allowed to use it?)

    Rapha

    Hi rapha,

    Yes that result rings a bell and I'm definitely allowed to use it. I seem to be a little rusty in analysis .

    Thanks for the super-quick response by the way!

    raheel88
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  4. #4
    MHF Contributor
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    Quote Originally Posted by raheel88 View Post
    Hi all,

    I'm trying toevaluate the following limit;

    limit as n --> oo (n-1)*(1/2)^(n-1)

    My intuition tells me that it should be zero because (1/2)^(n-1) approaches zero faster than (n-1) approaches infinity...but I'm not sure how to show this rigorously.

    Can anyone help?
    Yes, it goes to zero quite quickly.

    \lim_{n\rightarrow\infty}(n-1)\frac{1}{2^{n-1}}=\lim_{n\rightarrow\infty}\frac{n-1}{2^{n-1}}

    =2\lim_{n\rightarrow\infty}\frac{n}{2^n}-2\lim_{n\rightarrow\infty}\frac{1}{2^n}=0-0
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