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Math Help - limits of sequences

  1. #1
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    limits of sequences

    find the limit of each sequence:
    A) Sn = (1 + 1/2n)^(2n)
    B) Sn = ( 1 + 1/n)^(2n)
    C) Sn - (1 + 1/n) ^ (n-1)
    D) Sn = (n/(n+1))^n
    E) Sn = (1+ 1/2n)^n
    F) Sn = ((n+2)/(n+1))^(n+3)

    I found the answers to be this: A)e B) e^2 C) e D) 1/e E) sqrt(e) F) e...but I found the answers by using a different way then what the problems asks...it says to use the binomial theorem and the limit is referred to as e and is used as the base for natural logarithms.........Anyone know what they are asking and how to show the work needed to get the answers I got on A-F?
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  2. #2
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    Quote Originally Posted by learn18 View Post
    find the limit of each sequence:
    A) Sn = (1 + 1/2n)^(2n)
    B) Sn = ( 1 + 1/n)^(2n)
    C) Sn - (1 + 1/n) ^ (n-1)
    D) Sn = (n/(n+1))^n
    E) Sn = (1+ 1/2n)^n
    F) Sn = ((n+2)/(n+1))^(n+3)

    I found the answers to be this: A)e B) e^2 C) e D) 1/e E) sqrt(e) F) e...but I found the answers by using a different way then what the problems asks...it says to use the binomial theorem and the limit is referred to as e and is used as the base for natural logarithms.........Anyone know what they are asking and how to show the work needed to get the answers I got on A-F?
    This is a little bit informal but the explanation is clearer this way.

    A) By the binomial theorem:

    Sn = (1 + 1/2n)^(2n) = 1 + (2n) (1/(2n) + (2n)(2n-1)(1/2n)^2/2! + ..

    ...................................... (2n)!/[(2n-r)! r!] (1/(2n)^r + ..

    where r<=n.

    Now as n becomes large the term (2n)!/[(2n-r)! is the product of r factors
    of the form (2n-a) a = 0, 1, 2, .. r-1, which when divided by (2n)^r
    approaches 1. So the r-th term of Sn approaches 1/r! as n goes to infinity.

    Hence:

    Lim(n->infty) Sn = 1 + 1/1! + 1/2! + ... + 1/r! + ... = e.

    RonL
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  3. #3
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    Here are B and C they are baby stuff.
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  4. #4
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    ..
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  5. #5
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    Here is a last one and another attempt at #A.
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