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Math Help - Limit of Sequence

  1. #1
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    Limit of Sequence

    Prove that the limit as n approaches infinity of (2^n)/n! = 0.

    Any help?
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  2. #2
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    Have you considered the limit of the ratio [a_(n+1)/a_n]->0?
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  3. #3
    Member classicstrings's Avatar
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    Quote Originally Posted by JaysFan31 View Post
    Prove that the limit as n approaches infinity of (2^n)/n! = 0.

    Any help?

    Well just by looking at it, 2^a number is less than the same number factorial for large values.

    But that's no proof...more like logic...
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  4. #4
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    Quote Originally Posted by classicstrings View Post
    But that's no proof...more like logic...
    I am sure he realized that.

    He is probably taking a real analysis class he must post proofs. In calculus it is okay to say that but in a real analysis class you will fail if you say something like that.
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  5. #5
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    Maybe you can show me? Or not if you can't be bothered...
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    Maybe you can show me? Or not if you can't be bothered...
    Plato's hint is probably the best way:

    a_n = 2^n /n!

    So lim(n->inf) a_(n+1)/a_n = lin(n->inf)[2^(n+1)/(n+1)!]/[2^n/n!]

    = lim(n->inf)[2^(n+1)/2^n][n!/(n+1)!] = lim(n->inf)[2/(n+1)] = 0.

    Since this limit goes to zero, the limit of a_n goes to zero as n->infinity.

    -Dan
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  7. #7
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    Congratulations topsquark on your 1th post.
    ---
    The name of that rule is called the ratio test.
    If you ever studied infinite series that is what is used.
    ---
    I have very elegant explanation to why it works (again if you studied infinite series).

    You are given that sequence a_n
    Instead consider the sequence of partial sums s_n
    Since, by the ratio test lim s_n exists it implies that the originial sequence must satisfy lim a_n --> 0
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  8. #8
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    If {a_n+1/a_n}->0 and if 0<r<1 there is a positive integer K such that {a_K+1/a_K}<r.
    That implies that {a_K+1}<{a_K}r.
    But then {a_K+2/a_K+1}<r or {a_K+2}<{a_K+1}r<{a_K}r^2.
    We get for each n, {a_K+n}r<{a_K}r^n.
    Also because 0<r<1, (r^n)->0 so {a_K+n}->0.
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