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    Question - http://i.imgur.com/57uP0hZ.png
    My answer - http://i.imgur.com/czNTnqC.jpg

    1) Is what I've done correct?
    2) Could I have been quicker about it? (Our Professor does place a lot of emphasis on "rigor"...)

    Thanks
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    How do you get 1/|a_n| <A? Assuming you meant the same A as before...
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    (Original post by Enotdead)
    Question - http://i.imgur.com/57uP0hZ.png
    My answer - http://i.imgur.com/czNTnqC.jpg

    1) Is what I've done correct?
    2) Could I have been quicker about it? (Our Professor does place a lot of emphasis on "rigor"...)

    Thanks
    1. As pointed out, you've got the inequality wrong
    2. It's waffly rather than rigorous at the end.

    You need to choose an epsilon, then show that you can choose an appropriate "A" to make the inequality work. So something like:

    Let \epsilon &gt; 0.

    Then by the "which?" property of \mathbb{R}, there exists N \in \mathbb{N} such that "some number" < "some expression involving N and \epsilon".

    Since a_n \rightarrow \infty, we can find M \in \mathbb{N} such that

    n &gt; M \Rightarrow a_n &gt; N \Rightarrow \frac{1}{|a_n|} &lt; \frac{1}{N} &lt; \text{what?}

    I've left some "fill in the blanks" above for you to replace.
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    (Original post by Enotdead)
    Question - http://i.imgur.com/57uP0hZ.png
    My answer - http://i.imgur.com/czNTnqC.jpg

    1) Is what I've done correct?
    No. It's fatally flawed. You need to review your definitions, because the ones you're using are wrong.

    For example, the first 2 lines should have looked like:

    If a_n \to \infty then for any real number A > 0, we can find N \in \mathbb{N} such that a_n &gt; A for all n > N.

    Note the change in order. It's vitally important.
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    Thanks for the replies.
    Is this any better?
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    (Original post by Enotdead)
    Is this any better?
    Much better.

    The one small issue is that by stating the definition for a_n \to \infty first (before you start working on 1/a_n), it means that you haven't "really" defined N when you start using it later on.

    It works better to do something like:

    Take \epsilon &gt; 0. Since a_n \to \infty, we can find N such that a_n &gt; 1/\epsilon whenever n &gt; N. (and then carry on as in your proof).

    This makes it clear that the N we are now working with has been specifically chosen such that a_n &gt; 1/\epsilon.

    It doesn't actually cause much confusion here, but in a problem where you might have multiple sequences it makes more of a difference.
 
 
 
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