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Math Help - Elementary Analysis

  1. #1
    is up to his old tricks again! Jhevon's Avatar
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    Elementary Analysis

    So i got 8 questions for homework, i did 6. Here are the last two i'm having problems with. Thanks a lot guys.

    12.12
    Let (Sn) be a sequence of nonnegative numbers, and for each n, define
    An = (1/n)(s1 + s2 + ... + sn)

    (a) Show that liminfSn <= liminf An <= limsupAn <= limsupSn

    Hint: for the last inequality show first that M > N implies sup{An : n>M} <= (1/M)(s1 + s2 + ... + sN) + sup{Sn : n>N}

    (b) Show that if limSn exists, then limAn exists and limAn = limSn

    12.14
    Calculate:
    (a) lim(n!)^(1/n)
    (b) lim(1/n)(n!)^(1/n)

    So yea, that last problem seems like a calc 3 problem, but nevertheless, i dont know how to do it.

    Another thing that i probably should know already: Is adding sequences like adding vectors? as in you add corresponding terms of the sequences? What about multiplying two sequences, how is that done?

    I may come back and post my solutions for the questions i did just to see what you guys think, but only if i can find the time, this homework is due monday. thanks again
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  2. #2
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    Without TeX this next to impossible. Above I have done one case for you.
    Now let the sequence (S_n L) converges to zero. Use that to finish.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post


    Without TeX this next to impossible. Above I have done one case for you.
    Now let the sequence (S_n L) converges to zero. Use that to finish.

    this is for 1(b) right?
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  4. #4
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    Yes it is 1b.
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    Calculate:
    (a) lim(n!)^(1/n)
    Ah! These were the fun ones for homework.

    Let s_n=n!

    Note, that this is always non-zero.
    By the ratio test,
    lim |(n+1)!/n!|=lim (n+1) = +infinity.

    That means that, the root test always gives the same result.
    Hence,
    lim |n!|^{1/n}=+\infinity.
    (b) lim(1/n)(n!)^(1/n)
    Define a sequence of non-zero terms as,
    s_n=n!/(n^n)

    Then, by the ratio test, (details omitted)
    lim (n/(n+1))^n=1/lim(1+1/n)^n = 1/e.

    That means, the root test also gives the same result.
    Thus,
    lim |s_n|^(1/n)=1/n*(n!)^(1/n)=1/e

    -----------

    I did the proof Plato did differently from the hint in the book and differently from him.
    Since, LaTeX is disabled I cannot post it. Perhaps tomorrow I can show it you.
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  6. #6
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    Here is my proof of that problem.


    Note the inequalities #1 and #2 used in proof.
    #1 is trivial but #2 takes some work to show.

    I also use the fact that,
    (a+b)/(c+d)<= a/c+b/d for b,d>0

    I do not know if that makes sense to you.
    (EDIT. I just realized in the last line I forgot to write out the full sum. But I need to leave now cannot fix it, hopefully you understand what I am talking about).
    Attached Thumbnails Attached Thumbnails Elementary Analysis-picture9.gif  
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  7. #7
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    TPH. That is a nice proof.
    But, is it not the same as I gave?
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  8. #8
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    Quote Originally Posted by Plato View Post
    TPH. That is a nice proof.
    I would not say that it is no nice. Because it is a bit too long (especially when inequality #2 is proven).
    But, is it not the same as I gave?
    It has some resembelance. But my proof is not based on yours, rather that I how I written it.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jhevon View Post
    Another thing that i probably should know already: Is adding sequences like adding vectors? as in you add corresponding terms of the sequences? What about multiplying two sequences, how is that done?
    What about these questions?
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    What about these questions?
    Yes.

    The book implicity assumes that.

    But if {s_n} is a sequence and {t_n} is a sequence.
    (And I would imagine have the same domain. But it is standard for sequences to have the entire N domain).

    Then, {s_n+t_n} is a sequence {u_n} defined to
    u_n=s_n+t_n for all n.
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