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Math Help - analysis inequality

  1. #1
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    analysis inequality

    Would someone mind showing me how to do this proof?

    Use Bernoulli's inequality to show that (1 + (1/n))^n < (1 + (1/(n+1)))^(n+1) for all n in N.

    I'm given a hint of let x = -1/(n+1)^2.

    Thank you in advance.
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  2. #2
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     (1+x)^r \geq 1+rx


     \left(1 + \frac{1}{n}\right )^n < \left (1+ \frac{1}{n+1}\right )^{n+1}

     x = -\frac{1}{(n+1)^2}


     \left(1 - \frac{1}{(n+1)^2} \right)^n \geq 1-\frac{n}{(n+1)^2}
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  3. #3
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    Well what does this mean?

    How do you manipulate the inequality at the end?
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  4. #4
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    \left( {1 - \frac{1}{{n + 1}}} \right)^{n + 1} \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1}  = \left( {1 - \frac{1}{{\left( {n + 1} \right)^2 }}} \right)^{n + 1}

    \left( {1 - \frac{1}{{\left( {n + 1} \right)^2 }}} \right)^{n + 1}  \ge 1 + \left( {n + 1} \right)\left( {\frac{{ - 1}}{{\left( {n + 1} \right)^2 }}} \right) = \left( {1 - \frac{1}{{n + 1}}} \right)

    \left( {1 - \frac{1}{{n + 1}}} \right)^{n + 1} \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1}  \ge \left( {1 - \frac{1}{{n + 1}}} \right)

    \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1}  \ge \left( {1 - \frac{1}{{n + 1}}} \right)^{ - n}  = \left( {1 + \frac{1}{n}} \right)^n
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  5. #5
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    Is there a similar way to prove this inequality using the Arithmetic-Geometric Mean Inequality?

    Use x1 = 1, xk = (1 + (1/n)), and k = 2,3,...,n+1.
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  6. #6
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    Please review what you posted!
    It is not an inequality.
    In fact, it is hard to know what you may mean.
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  7. #7
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    Yeah, sorry for the confusion. I meant the original inequality (the one proven using Bernoulli's Inequality). This one:

    (1 + (1/n))^n < (1 + (1/(n+1)))^(n+1) for all n in N.


    This is just a hint for how to attack the problem as given in the textbook.
    Use x1 = 1, xk = (1 + (1/n)), and k = 2,3,...,n+1.
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  8. #8
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    The problem with hints is that often the only one who knows how it should be used is the person giving the hint. In this case, I have no idea.
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  9. #9
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    \displaystyle\frac{1+\left(1+\frac{1}{n}\right)+\l  eft(1+\frac{1}{n}\right)+\ldots+\left(1+\frac{1}{n  }\right)}{n+1}>\sqrt[n+1]{\left(1+\frac{1}{n}\right)^n}
    \displaystyle\frac{(n+1)+1}{n+1}>\left(1+\frac{1}{  n}\right)^{\frac{n}{n+1}}
    Rise both members to n+1.
    \left(1+\frac{1}{n+1}\right)^{n+1}>\left(1+\frac{1  }{n}\right)^n
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