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Math Help - Analysis problem on Sequences

  1. #1
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    Analysis problem on Sequences

    Hi guys, just got a question of Analysis on Sequences.

    a sequence e(n) (sry dun know how to type (en) little n.
    ok....

    e(n)=(1+1/n)^n

    (a). Show that e(n) is bounded above

    (b). Use The Monotone Convergence Theorem to conclude that e(n) converges to a limit "e", and give an interval of length less than one in which "e" lies. The number "e" is called the Euler Number.

    Appreciate!!!
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  2. #2
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    This may not be the neatest way to do this.

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

    binomial theorem
    = 1 + \frac{n}{n} + \frac{n(n-1)}{2 n^2} + \frac{n(n-1)(n-2)}{3! n^3} + \cdots +\frac{n!(n-1)(n-2) \cdots 3 \cdot 2\cdot 1}{n n^n}

     = 1 + 1 + \frac{1}{2} \left(1 - \frac{1}{n} \right) + \frac{1}{3!} \left(1 - \frac{1}{n} \right)  \left(1 - \frac{2}{n} \right) + \  \cdots   \ + \frac{1}{n!}  \left(1 - \frac{1}{n} \right) \cdots  \left(1 - \frac{n-1}{n} \right)

    bracketed terms are positive and all less than 1

    < 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}}

    less than infinite geometric sum \therefore  <3

    the rest should be easy.

    Bobak
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  3. #3
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    thx man
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  4. #4
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    but what about part (b).
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  5. #5
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    Well i showed you the limit is less than 3, can you show it is greater than 2 ?

    Bobak
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  6. #6
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    sure,thx man
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