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Math Help - two easy questions

  1. #1
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    two easy questions

    Here's something I thought would be easy to prove (if I'm corret of course)

    If x_n -> x and x >a, then x_n>a for infinitly many values of n.

    My working: we need |x_n -x|<|x-a|= x-a for n sufficently large

    ->  |x_n|<|x|+x-a

    whatever I do I can't seem to deduce x_n > a for all but finite n

    My second question is very simple. In the ratio test, if the limit does not exist what does limit superior greater than 1 imply?

    Thanks.
    Last edited by boromir; August 29th 2011 at 08:37 AM.
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  2. #2
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    Re: two easy questions

    Quote Originally Posted by boromir View Post
    Here's something I thought would be easy to prove (if I'm corret of course)
    If x_n -> x and x >a, then x_n>a for infinitly many values of n.
    Let \varepsilon  = \frac{{x - a}}{2} > 0

    If |x_n-x|<\varepsilon then a<\frac{a+x}{2}<x_n.
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  3. #3
    Senior Member vincisonfire's Avatar
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    Re: two easy questions

    Quote Originally Posted by boromir View Post
    My second question is very simple. In the ratio test, if the limit does not exist what does limit superior greater than 1 imply?
    If the inequality {a_{n+1}\over a_n} > 1 holds for all but finitely many n, then the series diverges. You can't say anything about limsup i think
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