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Thread: Prove that there is a threshold k such that xn>0 for all n>=k.

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    Prove that there is a threshold k such that xn>0 for all n>=k.

    Let (x_n) be a convergent sequence with limit a. Assume a>0. Prove that there is a threshold k such that x_n>0 for all n$\displaystyle \geqslant $k.

    What I have:

    Let $\displaystyle \varepsilon $>0. Then there is a k such that for all n$\displaystyle \geqslant $k, |x_n-a|<$\displaystyle \varepsilon $. Assume a>0. Then since $\displaystyle \exists $k such that ($\displaystyle \forall $n$\displaystyle \geqslant $k, |x_n-a|<$\displaystyle \varepsilon $. We know that 0<|[x_n-a|<$\displaystyle \varepsilon $ which is a<|xn|<$\displaystyle \varepsilon $+a. Since a>0 we have |xn|>a>0.
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    Quote Originally Posted by alice8675309 View Post
    Let (x_n) be a convergent sequence with limit a. Assume a>0. Prove that there is a threshold k such that x_n>0 for all n$\displaystyle \geqslant $k.
    Pick $\displaystyle \varepsilon = \frac{a}{2}$.
    If $\displaystyle |x_n-a|<\frac{a}{2}$ what do get when rewritten without absolute values signs?
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    Quote Originally Posted by Plato View Post
    Pick $\displaystyle \varepsilon = \frac{a}{2}$.
    If $\displaystyle |x_n-a|<\frac{a}{2}$ what do get when rewritten without absolute values signs?
    $\displaystyle \frac{a}{2}$<xn<$\displaystyle \frac{3a}{2}$ ? Which means that if $\displaystyle \frac{a}{2}$=$\displaystyle \varepsilon $ then xn>$\displaystyle \frac{a}{2}$=$\displaystyle \varepsilon $? so if xn>$\displaystyle \varepsilon $ and $\displaystyle \varepsilon $>0 so xn>0?
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    That does it. Right?
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    Quote Originally Posted by Plato View Post
    That does it. Right?
    yup. My question now is, when you look at a something to prove how do you know when to choose epsilon to be something other than just greater than 0 like you did in the problem by letting epsilon be a/2?
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    There is really no way to answer that question, other than to say one word: experience.
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