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Math Help - Limit Superior and Limit Inferior

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    Limit Superior and Limit Inferior

    I was just wondering if my method for finding lim sup and lim inf for sequences is correct. It's different to how we do it in class, but I asked my lecturer about my way and he said that it looks fine but he'd think about it and get back to me. Essentially what I do is first take the sequence to infinity and then find its sup and inf at infinity, instead of finding the sup/inf first and then taking this to infinity. Here is an example to clear up what I'm saying:

    Let the sequence x_n = (-1)^n + n/(n+1). What I first do is find the limit as n tends to infinity for each term. In this case, the first term alternates between 1 and -1, and the second term tends to 1. Once this is done I find the supremum and infimum. Therefore,

    lim sup x_n = 1 + 1 = 2 and lim inf x_n = -1 + 1 = 0.

    Is my method fine, or could there be a case where it would not work?
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  2. #2
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    Re: Limit Superior and Limit Inferior

    Quote Originally Posted by Alexrey View Post
    I was just wondering if my method for finding lim sup and lim inf for sequences is correct. It's different to how we do it in class, but I asked my lecturer about my way and he said that it looks fine but he'd think about it and get back to me. Essentially what I do is first take the sequence to infinity and then find its sup and inf at infinity, instead of finding the sup/inf first and then taking this to infinity. Here is an example to clear up what I'm saying:

    Let the sequence x_n = (-1)^n + n/(n+1). What I first do is find the limit as n tends to infinity for each term. In this case, the first term alternates between 1 and -1, and the second term tends to 1. Once this is done I find the supremum and infimum. Therefore,

    lim sup x_n = 1 + 1 = 2 and lim inf x_n = -1 + 1 = 0.

    Is my method fine, or could there be a case where it would not work?
    That method works fine, provided that one of the two sequences actually has a limit as n\to\infty. In the case x_n = (-1)^n + n/(n+1), the second sequence n/(n+1) tends to the limit 1, and the method works. But (to take a rather silly example) suppose that x_n = (-1)^n + (-1)^{n+1}. In that case, each of the two terms has supremum 1 and infimum 1, and your method would suggest that \limsup x_n = 2 and \liminf x_n = -2. But when you combine the two terms you see that x_n=0 for every n, and so \limsup x_n and \liminf x_n are both 0.
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