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    analysis - 2

    How does the comparison test incorporate the cauchy criterion for convergence?
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  2. #2
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    Quote Originally Posted by pila0688 View Post
    How does the comparison test incorporate the cauchy criterion for convergence?
    Let T_n=\sum_{n=1}^{N} c_n

    So the comparison test says that if |a_n|\leqslant c_n~{\color{red}\star} for all N\leqslant n with N\in\mathbb{N} and \sum c_n converges, then so does \sum a_n

    Proof: \sum c_n converges implies that T_n converges. Now since every convergent sequence is Cauchy \color{blue}\star it also follows that there exists some K that if K\leqslant N\leqslant M then \left|T_M-T_N\right|<\varepsilon. But because of how we defined T_N we see this is equivalent to \sum_{n=N}^{M}c_n<\varepsilon. But by \color{red}\star

    \sum_{n=N}^{M}|a_n|\leqslant\sum_{n=N}^{M}c_n<\var  epsilon

    And

    \left|\sum_{n=N}^{M}a_n\right|\leqslant\sum_{n=N}^  {M}|a_n|\leqslant\sum_{n=N}^{M}c_n<\varepsilon.

    From there we conclude that \sum a_n converges

    It is clear that we used the Cauchy convergence criterion at \color{blue}\star
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