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Math Help - Advanced Calc help

  1. #1
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    Advanced Calc help

    Three questions, hope that's not too many!

    1. I need to prove that the sequence
    a_n = 1 + \frac{1}{2} + \frac{1}{3} +...+\frac{1}{n}
    What I have is for any \varepsilon there exists m,n \geq n* such that |a_m - a_n|
    I take that as |(1 + \frac{1}{2} +...+\frac{1}{m}) - (1 + \frac{1}{2} +...+\frac{1}{n})|
    and get
    |\frac{1}{n + 1} + \frac{1}{n + 2} + ... + \frac{1}{m})|
    Here is my first question: when I put that into
    |\frac{1}{n + 1} + \frac{1}{n + 2} + ... + \frac{1}{m})| > \frac{1}{n}
    I need a fixed vaule for \varepsilon. What am I missing?


    2.
    Prove that every unbounded sequence contains a monotone subsequence that diverges to infinity.

    On this one, I know unbounded can mean divergent. What do I do next to prove this? I am confused on this one!



    3.
    Prove that if a sequence a_n converges to 0, and a sequence b_n is bounded, then the sequence a_n b_n converges to 0.

    I had gotten help on this before, but hadn't gotten very far. I'm starting with
    \left[ {\exists B > 0} \right]\left( {\left| {b_n } \right| < B} \right)\quad \& \quad \varepsilon  = \frac{\delta }{B}, but how would I put this into a proof for this? again, confused!



    Thanks in advance, and I might have some more later!
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  2. #2
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    Quote Originally Posted by harbong View Post
    Three questions, hope that's not too many!

    1. I need to prove that the sequence
    a_n = 1 + \frac{1}{2} + \frac{1}{3} +...+\frac{1}{n}
    What I have is for any \varepsilon there exists m,n \geq n* such that |a_m - a_n|
    I take that as |(1 + \frac{1}{2} +...+\frac{1}{m}) - (1 + \frac{1}{2} +...+\frac{1}{n})|
    and get
    |\frac{1}{n + 1} + \frac{1}{n + 2} + ... + \frac{1}{m})|
    Here is my first question: when I put that into
    |\frac{1}{n + 1} + \frac{1}{n + 2} + ... + \frac{1}{m})| > \frac{1}{n}
    I need a fixed vaule for \varepsilon. What am I missing?
    I see that you are using a Cauchy sequence. Perhaps you want to show it converges (which you did not mention in the post). But you cannot show that because the Harmonic series (the sum of reciprocals) diverges.
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  3. #3
    TD!
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    For question 1, you wrote "sequence" but you gave a (partial sum of a) series. Perhaps you meant:

    a_n  = 1,\frac{1}{2},\frac{1}{3}, \cdots ,\frac{1}{n}

    Now, do you want to show this is a Cauchy sequence? That is converges?
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  4. #4
    TD!
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    Quote Originally Posted by harbong View Post
    3.
    Prove that if a sequence a_n converges to 0, and a sequence b_n is bounded, then the sequence a_n b_n converges to 0.

    I had gotten help on this before, but hadn't gotten very far. I'm starting with
    \left[ {\exists B > 0} \right]\left( {\left| {b_n } \right| < B} \right)\quad \& \quad \varepsilon  = \frac{\delta }{B}, but how would I put this into a proof for this? again, confused!
    If you can prove that a majorant (upper bound) for this sequence converges, than this one converges as well:

    \left| {a_n b_n } \right| \le \left| {Ba_n } \right| = \left| B \right|\left| {a_n } \right| \to 0
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  5. #5
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    TD!:
    The sequence is the sum up to n.
    so,
    a_1 = 1
    a_2 = 1 + \frac{1}{2}
    a_3 = 1 + \frac{1}{2} + \frac{1}{3}
    ...
    a_n = 1 + \frac{1}{2}+ \frac{1}{3} + ... + \frac{1}{n}

    I think the sequence you described would be a_n = \frac{1}{n}
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    I see that you are using a Cauchy sequence. Perhaps you want to show it converges (which you did not mention in the post). But you cannot show that because the Harmonic series (the sum of reciprocals) diverges.
    sorry, I forgot to finish the instructions. You are correct, I need to prove it it cauchy. I didn't look at it as a harmonic series, that seems to be the solution. would it be appropiate to just state that the harmonic series diverges or do I need to go farther into it?
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  7. #7
    TD!
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    Okay, so you're looking at the partial sums of the sequence u(n) = 1/n.
    Since (we happen to know that) it diverges, the sequence won't be Cauchy.

    What is the assignment? You said: "I need to prove it it cauchy"...
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  8. #8
    TD!
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    Quote Originally Posted by harbong View Post
    2.
    Prove that every unbounded sequence contains a monotone subsequence that diverges to infinity.

    On this one, I know unbounded can mean divergent. What do I do next to prove this? I am confused on this one!
    Because the sequence u(n) is unbounded (I'll assume upwards without loss of generality), you can choose any M and find an index m so that u(m) > M. Choose an M with such a u(m) and take u(m) as first element of your subsequence. Since the sequence was unbounded, there will be an index p such that u(p) > u(m) (else, the sequence would have been bounded by u(m)). Take this one as second element so that you have a monotone sequence. Proceed with this argument, you are now constructing a monotone, divergent sequence.
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