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Math Help - Inequalities

  1. #1
    Senior Member Sampras's Avatar
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    Inequalities

    1. Prove that  n! > \left(\frac{n}{e} \right)^{n} for  n \geq 1 .

    Proof. We use induction on  n . For  n=1 ,  1 > \frac{1}{e} . Now suppose that for some  k > 1 ,  k! > \left(\frac{k}{e} \right)^{k} . So we want to show that  (k+1)! > \left(\frac{k+1}{e} \right)^{k+1} . Now  (k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k} . Can we somehow use the fact that  \left(1+ \frac{1}{n}\right)^{n} < e for all  n ? We could flip it so that  \frac{1}{e} < \frac{1}{\left(1+\frac{1}{n} \right)^{n}} for all  n . And so  \left(\frac{n}{e} \right)^{n} < \frac{n^n}{\left(1+\frac{1}{n} \right)^{n^2}} . Or another approach may be to show that  \frac{k^{k} \cdot (k+1)}{e^k} > \frac{(k+1)^{k+1}}{e^{k+1}} ? The denominator of the second term is larger, indicating that it would probably be smaller.

    2. Show that  n! < \left(\frac{n+1}{n} \right)^{n} for  n>1 , and deduce that  n! < e \left(\frac{n}{2} \right)^{n} .

    For this one you would use the AM-GM inequality? Because the LHS is  n(n-1)(n-2) \cdots 1 . So taking  \sqrt[n]{n!} we can use the AM-GM inequality?
    Last edited by Sampras; May 29th 2009 at 06:53 PM.
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  2. #2
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Sampras View Post
    1. Prove that  n! > \left(\frac{n}{e} \right)^{n} for  n \geq 1 .

    Proof. We use induction on  n . For  n=1 ,  1 > \frac{1}{e} . Now suppose that for some  k > 1 ,  k! > \left(\frac{k}{e} \right)^{k} . So we want to show that  (k+1)! > \left(\frac{k+1}{e} \right)^{k+1} . Now  (k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k} . Can we somehow use the fact that  \left(1+ \frac{1}{n}\right)^{n} < e for all  n ? We could flip it so that  \frac{1}{e} < \frac{1}{\left(1+\frac{1}{n} \right)^{n}} for all  n . And so  \left(\frac{n}{e} \right)^{n} < \frac{n}{\left(1+\frac{1}{n} \right)^{n^2}} . Or another approach may be to show that  \frac{k^{k} \cdot (k+1)}{e^k} > \frac{(k+1)^{k+1}}{e^{k+1}} ? The denominator of the second term is larger, indicating that it would probably be smaller.

    2. Show that  n! < \left(\frac{n+1}{n} \right)^{n} for  n>1 , and deduce that  n! < e \left(\frac{n}{2} \right)^{n} .

    For this one you would use the AM-GM inequality? Because the LHS is  n(n-1)(n-2) \cdots 1 . So taking  \sqrt[n]{n!} we can use the AM-GM inequality?
    For (1),  k^{k} \cdot (k+1) \not > (k+1)^{k+1} . So the inequality depends on the denominators  e^{k} and  e^{k+1} . Is this where we use the fact that  \left(1+\frac{1}{n} \right)^{n} < e for all  n ?
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  3. #3
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    Quote Originally Posted by Sampras View Post
    1. Prove that  n! > \left(\frac{n}{e} \right)^{n} for  n \geq 1 .

    Proof. We use induction on  n . For  n=1 ,  1 > \frac{1}{e} . Now suppose that for some  k > 1 ,  k! > \left(\frac{k}{e} \right)^{k} . So we want to show that  (k+1)! > \left(\frac{k+1}{e} \right)^{k+1} . Now  (k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k} .
    so we just need to prove that (k+1)\left(\frac{k}{e} \right)^k > \left(\frac{k+1}{e} \right)^{k+1}, which after simplifying gives us this equivalent inequality: \left(1 + \frac{1}{k} \right)^k < e, which we know is true.


    2. Show that  n! < \left(\frac{n+1}{n} \right)^{n} for  n>1.
    this is false! check what you wrote again!
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  4. #4
    Senior Member Sampras's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    so we just need to prove that (k+1)\left(\frac{k}{e} \right)^k > \left(\frac{k+1}{e} \right)^{k+1}, which after simplifying gives us this equivalent inequality: \left(1 + \frac{1}{k} \right)^k < e, which we know is true.



    this is false! check what you wrote again!
    It should be  n! < \left(\frac{n+1}{2} \right)^{n} . So basically you simplified the given inequality into an inequality we know?
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  5. #5
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    Quote Originally Posted by Sampras View Post
    It should be  n! < \left(\frac{n+1}{2} \right)^{n} .
    this is a very straightforward application of AM-GM: \frac{n+1}{2}=\frac{\frac{n(n+1)}{2}}{n}=\frac{1+2  + \cdots + n}{n} > \sqrt[n]{1 \cdot 2 \cdot \cdots n}=\sqrt[n]{n!}.
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  6. #6
    Senior Member Sampras's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    this is a very straightforward application of AM-GM: \frac{n+1}{2}=\frac{\frac{n(n+1)}{2}}{n}=\frac{1+2  + \cdots + n}{n} > \sqrt[n]{1 \cdot 2 \cdot \cdots n}=\sqrt[n]{n!}.
    And to deduce that  n! < e \left(\frac{n}{2} \right)^{n} we know that  n! < \left(\frac{n+1}{2} \right)^{n} . So maybe simplify the first inequality to get the second one? We can do the following:  n! < \left[\left(1+\frac{1}{n} \right) \cdot \frac{n}{2} \right]^{n} ? We then get the second inequality.

    Is this correct?
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  7. #7
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    Quote Originally Posted by Sampras View Post
    And to deduce that  n! < e \left(\frac{n}{2} \right)^{n} we know that  n! < \left(\frac{n+1}{2} \right)^{n} . So maybe simplify the first inequality to get the second one? We can do the following:  n! < \left[\left(1+\frac{1}{n} \right) \cdot \frac{n}{2} \right]^{n} ? We then get the second inequality.

    Is this correct?
    you just need to show that \left(\frac{n+1}{2} \right)^n < e \left(\frac{n}{2} \right)^n, which, again, after simplifying gives you this equivalent and true inequality: \left( 1+ \frac{1}{n} \right)^n < e.
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  8. #8
    Senior Member Sampras's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    you just need to show that \left(\frac{n+1}{2} \right)^n < e \left(\frac{n}{2} \right)^n, which, again, after simplifying gives you this equivalent and true inequality: \left( 1+ \frac{1}{n} \right)^n < e.

    But my simplification of the current inequality to the inequality we proved is incorrect?
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