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Thread: Inequalities

  1. #1
    Senior Member Sampras's Avatar
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    Inequalities

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

    Proof. We use induction on $\displaystyle n $. For $\displaystyle n=1 $, $\displaystyle 1 > \frac{1}{e} $. Now suppose that for some $\displaystyle k > 1 $, $\displaystyle k! > \left(\frac{k}{e} \right)^{k} $. So we want to show that $\displaystyle (k+1)! > \left(\frac{k+1}{e} \right)^{k+1} $. Now $\displaystyle (k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k} $. Can we somehow use the fact that $\displaystyle \left(1+ \frac{1}{n}\right)^{n} < e $ for all $\displaystyle n $? We could flip it so that $\displaystyle \frac{1}{e} < \frac{1}{\left(1+\frac{1}{n} \right)^{n}} $ for all $\displaystyle n $. And so $\displaystyle \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 $\displaystyle \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 $\displaystyle n! < \left(\frac{n+1}{n} \right)^{n} $ for $\displaystyle n>1 $, and deduce that $\displaystyle n! < e \left(\frac{n}{2} \right)^{n} $.

    For this one you would use the AM-GM inequality? Because the LHS is $\displaystyle n(n-1)(n-2) \cdots 1 $. So taking $\displaystyle \sqrt[n]{n!} $ we can use the AM-GM inequality?
    Last edited by Sampras; May 29th 2009 at 05:53 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 $\displaystyle n! > \left(\frac{n}{e} \right)^{n} $ for $\displaystyle n \geq 1 $.

    Proof. We use induction on $\displaystyle n $. For $\displaystyle n=1 $, $\displaystyle 1 > \frac{1}{e} $. Now suppose that for some $\displaystyle k > 1 $, $\displaystyle k! > \left(\frac{k}{e} \right)^{k} $. So we want to show that $\displaystyle (k+1)! > \left(\frac{k+1}{e} \right)^{k+1} $. Now $\displaystyle (k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k} $. Can we somehow use the fact that $\displaystyle \left(1+ \frac{1}{n}\right)^{n} < e $ for all $\displaystyle n $? We could flip it so that $\displaystyle \frac{1}{e} < \frac{1}{\left(1+\frac{1}{n} \right)^{n}} $ for all $\displaystyle n $. And so $\displaystyle \left(\frac{n}{e} \right)^{n} < \frac{n}{\left(1+\frac{1}{n} \right)^{n^2}} $. Or another approach may be to show that $\displaystyle \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 $\displaystyle n! < \left(\frac{n+1}{n} \right)^{n} $ for $\displaystyle n>1 $, and deduce that $\displaystyle n! < e \left(\frac{n}{2} \right)^{n} $.

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

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


    2. Show that $\displaystyle n! < \left(\frac{n+1}{n} \right)^{n} $ for $\displaystyle n>1.$
    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 $\displaystyle (k+1)\left(\frac{k}{e} \right)^k > \left(\frac{k+1}{e} \right)^{k+1},$ which after simplifying gives us this equivalent inequality: $\displaystyle \left(1 + \frac{1}{k} \right)^k < e,$ which we know is true.



    this is false! check what you wrote again!
    It should be $\displaystyle 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 $\displaystyle n! < \left(\frac{n+1}{2} \right)^{n} $.
    this is a very straightforward application of AM-GM: $\displaystyle \frac{n+1}{2}=\frac{\frac{n(n+1)}{2}}{n}=\frac{1+2 + \cdots + n}{n} > \sqrt[n]{1 \cdot 2 \cdot \cdots n}=\sqrt[n]{n!}.$
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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: $\displaystyle \frac{n+1}{2}=\frac{\frac{n(n+1)}{2}}{n}=\frac{1+2 + \cdots + n}{n} > \sqrt[n]{1 \cdot 2 \cdot \cdots n}=\sqrt[n]{n!}.$
    And to deduce that $\displaystyle n! < e \left(\frac{n}{2} \right)^{n} $ we know that $\displaystyle n! < \left(\frac{n+1}{2} \right)^{n} $. So maybe simplify the first inequality to get the second one? We can do the following: $\displaystyle 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 $\displaystyle n! < e \left(\frac{n}{2} \right)^{n} $ we know that $\displaystyle n! < \left(\frac{n+1}{2} \right)^{n} $. So maybe simplify the first inequality to get the second one? We can do the following: $\displaystyle 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 $\displaystyle \left(\frac{n+1}{2} \right)^n < e \left(\frac{n}{2} \right)^n,$ which, again, after simplifying gives you this equivalent and true inequality: $\displaystyle \left( 1+ \frac{1}{n} \right)^n < e.$
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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 $\displaystyle \left(\frac{n+1}{2} \right)^n < e \left(\frac{n}{2} \right)^n,$ which, again, after simplifying gives you this equivalent and true inequality: $\displaystyle \left( 1+ \frac{1}{n} \right)^n < e.$

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