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Math Help - Prove by contradiction that the sequence is divergent

  1. #1
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    Prove by contradiction that the sequence is divergent

    This is a tough one, and I have little idea on how to prove it.

    Prove that the sequence defined by f(n)=(-1)^n,n\in N is divergent.

    This must be proven by contradiction.
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  2. #2
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    Quote Originally Posted by Runty View Post
    This is a tough one, and I have little idea on how to prove it.

    Prove that the sequence defined by f(n)=(-1)^n,n\in N is divergent.

    This must be proven by contradiction.

    Suppose (-1)^n\xrightarrow [n\to\infty]{}a , for some a\in\mathbb{R}\Longrightarrow \forall\,\epsilon>0\,\,\,\exists\,N_\epsilon\in\ma  thbb{N}\,\,\,s.t.\,\,\,n>N_\epsilon\Longrightarrow  \,|(-1)^n-a|<\epsilon .

    Let us choose \epsilon=\frac{1}{4}\Longrightarrow for some natural number N_\frac{1}{4} , we have that n>N_\frac{1}{4}\Longrightarrow |(-1)^n-a|<\frac{1}{4} \Longleftrightarrow a-\frac{1}{4}<(-1)^n<a+\frac{1}{4} .

    But this means that all but a finite number of the sequence's elements are between the left and the right ends, and the difference between these two ends is a+\frac{1}{4}-(a-\frac{1}{4})=\frac{1}{2} , and this is a contradiction since for any two consecutive natural numbers n_1,n_2>N_\frac{1}{4}\,,\,\,|(-1)^{n_1}-(-1)^{n_2}|=2 , so it's impossible that ALL the elements of the seq. (but perhaps a finite number of them) are within two numbers whose difference is 0.5 ...

    Tonio
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  3. #3
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    Also, an alternative way would be to notice that if the sequence is convergent then it is cauchy, but for each n \in \mathbb{N}, \ |a_{n+1} - a_n| =2
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