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Math Help - Some questions on proving convergence tests

  1. #1
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    Some questions on proving convergence tests

    I was reading proofs on some convergence tests, but I'm stuck on a few parts here:

    1. Prove every cauchy sequence is bounded.

    Proof: Suppose the sequence {An} is cauchy.
    Let E = 1, pick N in Natural Numbers such that n >= N and m >= N

    We have |An - Am| < 1, further, we have |An-AN| < 1.

    This is the part that I don't understand, how do we know that|An - AN| < 1?

    2. For a number r such that |r| < 1, then the series sum {from k=0} r^k = 1/(1-r).

    Proof: sum {from k=0 to n} r^k = 1 + r + r^2 + r^3 + . . . + r^n = (1 - r^{n+1}) / (1 - r)

    Now, how do we know that equals?

    Thank you, y'all!

    KK
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  2. #2
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    Theorem: If (s_n) is a Weierstrauss sequence then s_n is bounded.

    Proof: We now that for any \epsilon we can choose N\in \mathbb{N} so that n,m>N\implies |s_n-s_m|<\epsilon. So choose \epsilon = 1 and we have |s_n - s_m|<1. That means
    |s_n|-|s_m|\leq|s_n - s_m|<1 \mbox{ thus }|s_n|< 1+ |s_{N+1}| (by choose m=N+1).
    Now the above inequality if true for only n>N.
    So,
    |s_n|<1+|s_{N+1}| for n>N.
    But what if n=1,2,..,N?
    No problem, let,
    M= \max \{ s_1,s_2,...,s_n , 1+|s_{N+1}| \}.
    And so,
    |s_n|\leq M \mbox{ for all } n\in \mathbb{N}.
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    2. For a number r such that |r| < 1, then the series sum {from k=0} r^k = 1/(1-r).
    If |r|<1 then,
    1+r+r^2+...+r^n = \frac{1-r^{n+1}}{1-r}\to \frac{1}{1-r} \mbox{ as }n\to \infty \mbox{ since }|r|<1
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  4. #4
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    I understand the first proof now, thank you.

    But in the second one, why does

    <br />
1+r+r^2+...+r^n = \frac{1-r^{n+1}}{1-r}<br />

    true?

    Other than that, I understand the rest, thank you!
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  5. #5
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    Geometric series formula + the following theorem:
    \lim \ a^n = 0 \mbox{ if } |a|<1.
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