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Sequence question. Need help

  1. Apr 22, 2007 #1
    sequence question. Need help!!

    i would like to know where could i find the proof that the sequence

    a_n=(1+1/n)^n bounded (upper bounded) by 4.

    or in general that this sequence is a convergent one??

    i know the proof by expanding it using binominal formula(Newton formula), but i am looking for another proof, by using some other helping sequence, and than to tell that this sequence a_n=(1+1/n)^n is smaller than every term of the helpin sequence????

    any help would be appreciated.
  2. jcsd
  3. Apr 22, 2007 #2


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    Have you looked at the power series for e?
  4. Apr 23, 2007 #3
    do u mean expressing e using taylor formula??
  5. Apr 23, 2007 #4

    Gib Z

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    I Think thats what he meant, but heres a definition of e that will help :)

    [tex]e=\lim_{n\to\infty} (1+\frac{1}{n})^n[/tex].
    So as n goes to infinity, a_n goes to e, which is less than 4.

    However, you need to know that a_n < 4 for ANY n. You know a_1=2.

    So to make sure for any positive integer n, 2<a_n<4, we show that that a_n is an monotonically increasing function for positive n.
    Last edited: Apr 23, 2007
  6. Apr 23, 2007 #5
    Yes, i do know this. But what i am looking for is a proof(another proof, couse i already know two of them) using another sequence that will look something like this


    and than to show that for every n, a_n, is less than b_n, for every n.(a_n<b_n)

    because i know that 2<e<3<4 .

    because i need to prove it this whay(using another helping sequence which charasteristics we know).

    thnx anywhay
  7. Apr 23, 2007 #6

    Gib Z

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    Well I'm not sure about the form of your b_n, but heres another b_n thats larger term by term and is bounded by 4 as well.

    [tex]e^x = \lim_{n\to\infty} (1+\frac{x}{n})^n[/tex]
    So a b_n that is larger term by term and bounded we be say..b_n = (1 + 1.002/n)^n? Any value of or less than ln 4 will do in place of 1.002.
  8. Apr 23, 2007 #7


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    e=1+1/1!+1/2!+1/3!+.... which dominates the binomial expansion of (1+1/n)n
  9. Apr 24, 2007 #8

    How does this help, to prove what i am looking for???
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