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Homework Help: Euler's number e, proving convergence and bounds

  1. Oct 22, 2005 #1
    The transcendental number e – Euler’s number
    (the underscores(_) represent subscript and (^) represent superscript/exponet)
    The limiting value of sequence {e_n} where e_n = (1+1/n)^n is the irrational number e. My text gives a challenge to see if you can prove that this converges by verifying the following: (1+1/n) ^n < e < (1+1/n) ^(n+1) and that 2 ≤ e < 3 for all n Є N. There has been a consensus in the class that there is a typo in the text and that it should read: (1+1/n) ^n < e_(n+1) < (1+1/n)^( n+1) 2 ≤ e_m < 3. I’m not sure if I know enough to question this but I will accept it and proceed with the hints on how to do this.

    The first hint is to
    Use the binominal theorem to write out n+1 terms of e_n using the fact that n(n-1)(n-2)/n^3 can be expressed as (1-1/n)(1-2/n). then do the same for n+2 in terms of e_(n+1) such that you can come to the conclusion that e_n < e_(n+1) right off the bat in this multi step problem I’m stumped am I using the binominal expansion correctly?. So far I’ve come up with:

    e_n = 1 + 1 + (1/2)(n)(n-1)(1/n^2) + (1/3)(n)(n-1)(n-2)(1/n^3) + …. + (1/n)^n
    e_n = 2 + (1/2)(n)(n-1)(1/n^2) + (1/3)(1-(1/n))(1-(2/n)) + …. + (1/n)^n

    then for n+1 in terms of e_n

    = 2 + (1/2)(n+1)(n)(1/(n+1)^2) + (1/3)(1-(1/(n+1))(1-2/(n+1)) + …. + (1/(n+1))^(n+1)

    Then for n+2 in terms of e_(n+1) here I substituted n+2 for n in the equation above, is this correct according to the given hints?

    = 2 + (1/2)(n+3)(n+2)(1/(n+3)^2) + (1/3)(1-(1/(n+3))(1-2/(n+3)) + …. + (1/(n+3))^(n+3)

    I wanted to get some help on these preliminary steps before I move onto the second half of the challenge. If I am correct so far how does this come to the conclusion that e_n < e_(n+1)?
     
    Last edited: Oct 22, 2005
  2. jcsd
  3. Oct 22, 2005 #2

    VietDao29

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    Oh, it's very hard to read... You can learn how to LaTeX at the General Physics board.
    Nope that's not quite correct, you cannot substitute n + 1 or n + 2 for n, note that en has n + 1 terms while en + 1 has n + 2 terms, and en + 2 has n + 3 terms. You should again use binomial theorwm to write out the terms for en + 1, and en + 2

    You seem to forget the factorial sign in while expanding the terms out :wink:...
    It should read something like:
    [tex](a + b) ^ n = a ^ n + na ^ {n - 1}b + \frac{n(n - 1)}{2}a ^ {n - 2}b ^ 2 + \frac{n(n - 1)(n - 2)}{3!}a ^ {n - 3}b ^ 3 + ... + b ^ n[/tex]

    It seem that you forget to rewrite your third terms for en. It can be written as:
    [tex]\frac{1}{2} \times \frac{n(n - 1)}{n ^ 2} = \frac{1}{2} \times \frac{n}{n} \times \frac{n - 1}{n} = \frac{1}{2} \times \left( 1 - \frac{1}{n} \right)[/tex]

    Then you can continue your proof by noting that:
    [tex]\frac{m}{n} > \frac{m}{n + 1}, \forall n \geq 1, m > 0[/tex]
    Hence
    [tex]1 - \frac{m}{n} < 1 - \frac{m}{n + 1}, \forall n \geq 1, m > 0[/tex]

    Can you go from here?
    Viet Dao,
     
    Last edited: Oct 22, 2005
  4. Oct 23, 2005 #3
    Thanks for the latex info, i was wondering how to enact the various notations.
    Before I continue with the proof I want to verify that I'm using the binominal theorem correctly as well as latex.

    [tex]e_{n+1}=2+\frac{n-1}{2n}+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+....+(\frac{1}{n})^n+(\frac{1}{n+1})^{n+1}[/tex]

    [tex]e_{n+2}=2+\frac{n-1}{2n}+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+....+(\frac{1}{n})^n+(\frac{1}{n+1})^{n+1}+(\frac{1}{n+2}^{n+2})[/tex]


    I feel like i'm missing something, but anyway if I'm correct on the above expansion it seems that it would follow that [tex]e_{n}<e_{n+1}[/tex] since there is an additional expansion term that is >0 for all [tex]n\in{N}[/tex]

    I do follow your the latter part of how to do the proof, however my text also throws in that i should use this additional inequality [tex]n^{n}\geq{n!}\geq{2}^{n-1}[/tex] to prove that [tex]e_{n}<3[/tex]. if I get that, from here i can use the squeez theroem to prove that e is bounded between 2 and 3. I guess I should accept the given inequality to be true, but i thought that factorials grow larger faster than exponets. Thanks again for the help:smile:
     
    Last edited: Oct 23, 2005
  5. Oct 23, 2005 #4

    VietDao29

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    Yes, you do miss something. I think you should check the expansion for en + 1, and en + 2 again:
    [tex]e_{n + 1} = 1 + \frac{n + 1}{n + 1} + \frac{1}{2} \times \frac{(n + 1)n}{(n + 1) ^ 2} + \frac{1}{3!} \times \frac{(n + 1)n(n - 1)}{(n + 1) ^ 3} + ... + \frac{1}{(n + 1)!} \times \frac{(n + 1)n(n - 1) ... 1}{(n + 1) ^ {n + 1}}[/tex]
    [tex]= 2 + \frac{1}{2} \times \frac{n}{(n + 1)} + \frac{1}{3!} \times \frac{n(n - 1)}{(n + 1) ^ 2} + ... + \frac{1}{(n + 1)!} \times \frac{n(n - 1) ... 1}{(n + 1) ^ {n}}[/tex]

    [tex]= 2 + \frac{1}{2} \times \left( \frac{n + 1 - 1}{(n + 1)} \right) + \frac{1}{3!} \left( \frac{(n + 1 - 1)(n + 1 - 2)}{(n + 1) ^ 2} \right)[/tex]
    [tex]+ ... + \frac{1}{(n + 1)!} \times \frac{(n + 1 - 1)(n + 1 - 2)(n - + 1 - 3)...(n + 1 - n)}{(n + 1) ^ {n}}[/tex]

    [tex]= 2 + \frac{1}{2} \times \left( 1 - \frac{1}{(n + 1)} \right) + \frac{1}{3!} \left( 1 - \frac{1}{(n + 1)} \right) \times \left( 1 - \frac{2}{(n + 1)} \right) + ... +[/tex]
    [tex]+ \frac{1}{(n + 1)!} \left( 1 - \frac{1}{n + 1} \right) \left( 1 - \frac{2}{n + 1} \right) \left( 1 - \frac{3}{n + 1} \right)...\left( 1 - \frac{n}{n + 1} \right)[/tex]
    Now, you can use the last part of my previous post to prove that en < en + 1

    Of course, it's obvious that en > 2.
    You can also prove the inequality n! >= 2n - 1:
    n! = 1 . 2 . 3 . 4 . 5 . ... . n = 2 . 3 . 4 . ... . n >= 2 . 2 . 2 . 2 . ... . 2 = 2n - 1 Q.E.D
    To continue your proof, you should note that:
    [tex]1 - \frac{1}{m} < 1 , \forall m > 0[/tex]
    and [tex]\frac{1}{n!} \leq \frac{1}{2 ^ {n - 1}} , \forall n > 0[/tex]
    Can you go from here?
    Viet Dao,
     
    Last edited: Oct 23, 2005
  6. Oct 24, 2005 #5
    wow... I think I need further instruction on the binominal expansion. I'm a bit confused (not in the algebra afterwards, but how the part with the 3! term:
    [tex]\frac{1}{3!} \times \frac{(n + 1)n(n - 1)}{(n + 1) ^ 3}[/tex]

    this is probably due to my misunderstanding of this theorem but why is there (n+1)! in the last term
    [tex]\frac{1}{(n + 1)!} \times \frac{(n + 1)n(n - 1) ... 1}{(n + 1) ^ {n + 1}}[/tex]

    when in the [tex]e_{n}[/tex] last term there isn't the factorial?
     
  7. Oct 24, 2005 #6

    VietDao29

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    You should re-read your book on binomial expansion. Or you can also read it here (pay attention to the I part).
    It says:
    [tex](x + y) ^ n = \sum_{k = 0} ^ n \left( \begin{array}{l} n & k \end{array} \right) x ^ k y ^ {n - k}[/tex]
    Where:
    [tex]\left( \begin{array}{l} n & k \end{array} \right) = \frac{n!}{k!(n - k)!} = \frac{1}{k!} \times (n - k + 1)(n - k + 2) ... n[/tex]
    So let x = 1 / n, and y = 1.
    The forth term of the expansion for en + 1 will be:
    [tex]\left( \begin{array}{c} n + 1 & 3 \end{array} \right) \frac{1}{(n + 1) ^ 3} \times 1 ^ {n + 1 - 3} = \frac{(n + 1)!}{3!(n - 2)!} \times \frac{1}{(n + 1) ^ 3} \times 1 ^ {n - 2}[/tex]
    [tex]= \frac{1}{3!} (n - 1)n(n + 1) \times \frac{1}{(n + 1) ^ 3} \times 1 ^ {n + 1 - 3} = \frac{1}{3!} \times \frac{(n + 1)n(n - 1)}{(n + 1) ^ 3}[/tex].
    So do you get it now?
    For the last term, you may write it as:
    [tex]\frac{1}{(n + 1) ^ {n + 1}}[/tex]
    But writing it as:
    [tex]\frac{1}{(n + 1)!} \times \frac{(n + 1)n(n - 1) ... 1}{(n + 1) ^ {n + 1}}[/tex] makes it easier for you to prove en + 1 > en
    They are the same:
    [tex]\frac{1}{(n + 1)!} \times \frac{(n + 1)n(n - 1) ... 1}{(n + 1) ^ {n + 1}} = \frac{1}{(n + 1)!} \times \frac{(n + 1)!}{(n + 1) ^ {n + 1}} = \frac{1}{(n + 1) ^ {n + 1}}[/tex].
    Viet Dao,
     
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