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Math Help - Proof analysis

  1. #1
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    Proof analysis

    If {a_n} is a null sequence, and for each n, b_n is the arithmetic mean of the first n terms of {a_n}, then prove {b_n} is a null sequence.
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  2. #2
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    Quote Originally Posted by natester View Post
    If {a_n} is a null sequence, and for each n, b_n is the arithmetic mean of the first n terms of {a_n}, then prove {b_n} is a null sequence.
    b_N=\frac{1}{N}~\sum_{n=1}^N a_n

    Put M=\lfloor \sqrt{N} \rfloor then:

    b_N=\frac{1}{N}~\left[\sum_{n=1}^M a_n+\sum_{n=M+1}^N a_n\right] <br />
\le \frac{{\rm{sup}}_{n\in [1,M]}(a_n)}{\sqrt{N}}+\frac{N-(\sqrt{N}-1)}{N}{\rm{sup}}_{n \in [M,N]}(a_n)<br />

    The first term on the right clearly goes to 0 as N goes to \infty.

    The coefficient of the second term has a finite limit as N goes to \infty, but because \{a_n\} is a null sequence:

    \lim_{N \to \infty}{\rm{sup}}_{n \in [M,N]}(a_n)=0

    which proves that \{b_n\} is a null sequence

    RonL
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  3. #3
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    If fact is a_n \to a then b_n \to a too. Where b_n is an arithmetic sequence.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    If fact is a_n \to a then b_n \to a too. Where b_n is an arithmetic sequence.
    Which is a corrolary of what the original poster was asked to prove

    RonL
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