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Math Help - Sequence question

  1. #1
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    Sequence question

    I am studying for my final in undergrad analysis, and have a question concerning sequences. Here it is. Let \{s_n\} be a sequence \backepsilon \{s_n\}>0 \; \forall \; n \in \mathbb{N}. Suppose  \lim_{n \to \infty}s_n=s. Show that s \ge 0.

    Thanks in advance.
    -the Doctor
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  2. #2
    Senior Member jakncoke's Avatar
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    Pf: Assume  s_n > 0 \forall n \in \mathbb{N} and  \displaystyle\lim_{n\to\infty} s_n = s Now assume  s < 0 So, By def \forall \epsilon > 0 \exists N \in \mathbb{N} | \forall n \geq N \: |s_n - s|<\epsilon By the triangle ineq  \forall \epsilon > 0 \: |s_n| - |s| < \epsilon which means  |s_n| \rightarrow |s| as n \rightarrow \infty since  s_n > 0 \forall n we have  s_n \rightarrow |s| as n \rightarrow \infty since  |s|>0 this is a contradiction.
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  3. #3
    Member mohammadfawaz's Avatar
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    In other words, if s<0, then for a given \epsilon with s+\epsilon<0, you should have elements of s_n existing in the \epsilon neighborhood of s (by limit to s). Since s+\epsilon<0, then those elements are also negative which gives a contradiction.
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  4. #4
    Senior Member Pinkk's Avatar
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    I think there's a somewhat easier way. Suppose that s_{n} > 0 for all n\in \mathbb{N}. Suppose \lim_{n\to \infty} s_{n} = s < 0. Then we know there exists N such that n>N implies |s_{n} - s| < -s. This is true because such an implication must hold for any \epsilon > 0 and if s < 0 then -s > 0, so this is the particular case where \epsilon = -s. So this means s < s_{n} - s < -s for n>N. Adding s to each part of the equality, we get for n>N, 2s < s_{n} < 0. That means after some point, s_{n} will be negative! And there arises your contradiction.
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