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Math Help - Inequality

  1. #1
    Member great_math's Avatar
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    Inequality

    prove that for any n

    2\le\left(1+\frac{1}{n}\right)^n<3
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  2. #2
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    Do you know BERNOULLI's INEQUALITY? x >  - 1 \Rightarrow \quad \left( {1 + x} \right)^n  \geqslant 1 + nx.
    So \left( {1 + \frac{1}{n}} \right)^n  \geqslant 1 + n\left( {\frac{1}{n}} \right) = 2.
    That gives one ‘side’ of the inequality.

    Using Napper inequality we can get \ln \left( {1 + \frac{1}{n}} \right) < \frac{1}{n}.
    Using the exponential we get \left( {1 + \frac{1}{n}} \right) < e^{\frac{1}<br />
{n}}  \Rightarrow \left( {1 + \frac{1}{n}} \right)^n  < e < 3
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  3. #3
    Jes
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    What a slick way to get the lower bound. I normally induct using the binomial expansion or the identity a^n-b^n = (b^n + b^{n-2}a + \cdots + a^{n-1}) to derive b^n < a^n + (b-a)nb^{n-1} in order to show the sequence is monotonic increasing. Then it follows that \inf \left (n + \frac{1}{n}\right )^n = 2 since it is the first term. Let's see if I can use your method.

    Let a_n denote the sequence and let M = \inf (a_n). Assume M is such that 2 < M \leq \left (n + \frac{1}{n}\right )^n . Then a_1 = 2 < M contradicts that M is a lower bound.

    Next assume M < 2. There is some a_i in the sequence such that M \leq a_i < 2 otherwise 2 is the greatest lower bound. But no such a_i exists since it was proved that 2 \leq a_n < 3. By Trichotomy, \inf (a_n) = 2.
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