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Math Help - Is there a way to show my work for this limit?

  1. #1
    s3a
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    Is there a way to show my work for this limit?

    Is there a way to show my work for this? (other than knowing that both will grow infinitely and by pluggin in numbers to see which one grows faster): http://www.wolframalpha.com/input/?i=lim+as+n-%3Einf+((n!)%2F(2^n))

    Any input would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    You can easily show by induction that for n>6, n! > 3^n. Then \lim_{n \to \infty}\frac{n!}{2^n}>\lim_{n \to \infty}\frac{3^n}{2^n}=\infty.

    Post again in this thread if you are still having trouble.
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  3. #3
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    Hello, s3a!

    Converge or diverge? . \lim_{n\to\infty}\frac{n!}{2^n}
    Did you try the Ratio Test?


    . . R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!}{2^{n+1}}\cdot\frac{2^n}{n!} \;=\;\frac{(n+1)!}{n!}\cdot\frac{2^n}{2^{n+1}} \;=\;\frac{n+1}{2}


    Then: . \lim_{n\to\infty} R \;=\;\lim_{n\to\infty}\frac{n+1}{2} \;=\;\infty


    . . The sequence diverges.

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  4. #4
    s3a
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    What's the induction method?
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    I'll start by saying I like Soroban's answer. It's more clear and direct. However, if you don't know about the ratio test...

    By induction, I mean the method of proof where you prove some statement P(n) for n=1, then prove that P(n+1) follows from P(n). You can then conclude that P(n) is true for all natural numbers.

    So, if n=7, n!=5040 and 3^n=2187, so in this case n! > 3^n.

    Assuming that for some n, n! > 3^n, since n > 6 > 2, n+1 > 3 and we can multiply the inequalities, giving (n+1)! > 3^(n+1). (Note that the ratio test is hidden in the previous line of reasoning).

    So n! > 3^n for all n > 6. Of course, I chose n > 6 because 7 is the first integer for which it is true: 6! = 720 is not greater that 3^6 = 729.

    I hope that helps. Go ahead and post again in this thread if you're still having trouble.

    - Hollywood
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