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Math Help - Converging Sequences

  1. #1
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    Converging Sequences

    Ok.I cant think how to start either of these problems.

    For each sequence, determine whether the sequence converges and if it does, find the limit.

    a) (e^(npii))/1+3^n

    b) sinh[(npii)/2]


    a) I presuming that the dominant term is e, but when I divide by it I get

    1/[(1/e^npii)+(3^n)/e^npii]

    Is this the right way of doing it? If not can someone point me in the right direction.


    b)For this one I had

    sinh[(npii)/2]=(e^(npii/2) - e^(-npii/2))

    I tired a way using the the triangle inequality followed by the squeeze rule, but that came out wrong, so I dont know where I went wrong.

    If anyone could please, please help, I would be very grateful.

    Thanx

    Bex
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  2. #2
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    Quote Originally Posted by bex23 View Post
    Ok.I cant think how to start either of these problems.

    For each sequence, determine whether the sequence converges and if it does, find the limit.

    a) (e^(npii))/1+3^n
    Can we assume that this is (e^(npi i)}/(1+ 3^n)?
    Is so, that's east: |e^(npi i)|<= 1 while the denominator increases without limit.

    b) sinh[(npii)/2]
    sinh(z)= (e^(z)- e^(-z))/2i so sinh(npi i)/2= [(e^(n pi i)- e^(n pi i))/2] (1/2i)= sin(n pi) (-i/2) and sin(n pi)= 0 for all n.

    a) I presuming that the dominant term is e, but when I divide by it I get

    1/[(1/e^npii)+(3^n)/e^npii]

    Is this the right way of doing it? If not can someone point me in the right direction.


    b)For this one I had

    sinh[(npii)/2]=(e^(npii/2) - e^(-npii/2))

    I tired a way using the the triangle inequality followed by the squeeze rule, but that came out wrong, so I dont know where I went wrong.

    If anyone could please, please help, I would be very grateful.

    Thanx

    Bex
    Follow Math Help Forum on Facebook and Google+

  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Can we assume that this is (e^(npi i)}/(1+ 3^n)?
    Is so, that's east: |e^(npi i)|<= 1 while the denominator increases without limit..
    So this means the sequence is divergent right?


    Quote Originally Posted by HallsofIvy View Post
    sinh(z)= (e^(z)- e^(-z))/2i so sinh(npi i)/2= [(e^(n pi i)- e^(n pi i))/2] (1/2i)= sin(n pi) (-i/2) and sin(n pi)= 0 for all n.
    Not sure that this is correct. But isnt sinh(z)= (e^(z)- e^(-z))/2

    The formula I have does not include i. That would be sin(z) not sinh(z). Also the values seem to have got mixed up. It should be

    sinh[(npii)/2]

    Hopefully you can help me out.

    Thanx
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  4. #4
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by bex23 View Post
    So this means the sequence is divergent right?
    It means it's convergent, it converges to 0.
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  5. #5
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    Quote Originally Posted by Showcase_22 View Post
    It means it's convergent, it converges to 0.
    But isnt

    e^(pii)=-1 so

    e^(npii)=(e^pii)^n= (-1)^n

    and so is divergent
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  6. #6
    Super Member Showcase_22's Avatar
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    Factor in the denominator and it converges to 0.....

    Am I looking at the right sequence?
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  7. #7
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    Quote Originally Posted by Showcase_22 View Post
    Factor in the denominator and it converges to 0.....
    Can you maybe explain what you mean. I am having a slow day today, lol


    Quote Originally Posted by Showcase_22 View Post
    Am I looking at the right sequence?
    The sequence is the part (a) one. That is

    [e^(npii)]/(1+3^n)
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  8. #8
    Super Member Showcase_22's Avatar
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    Isn't a_n=\frac{(-1)^n}{1+3^n}?
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