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Math Help - Critique my proof?

  1. #1
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    [Solved] ....

    Claim: The series \sum_{n=1}^{ \infty} \frac{2}{n^2 + n} converges.

    By definition:
    The series \sum_{n=1}^{ \infty} \frac{2}{n^2 + n} = \sum_{n=1}^{ \infty} a_n converges if and only if for every \epsilon > 0, there exists some N \in \mathbb{N} such that for all n, m with n > m \geq N, we have |a_{m+1} + a_{m+2}+ ...+ a_n| < \epsilon.

    (Attempted) proof:
    Let m \in \mathbb{N} be arbitrary but fixed, and let S_m = \sum_{n=1}^{m} \frac{2}{n^2 + n}. Then the sequence (S_m) is convergent, and it necessarily follows that (S_m) is Cauchy. That is, for all \epsilon > 0, there exists some N \in \mathbb{N} such that for all n, m \geq N, we have |S_{n} - S_{m}| < \epsilon.

    But if n > m \geq N, we have |S_n - S_m| = |a_{m+1} + a_{m+2}+ ...+ a_n| < \epsilon

    which shows, directly from the definition, that the series \sum_{n=1}^{ \infty} \frac{2}{n^2 + n} is convergent. \square

    Is this satisfactory? Or is there a more simplistic approach that I should be taking?
    Last edited by CaptainBlack; August 24th 2010 at 02:24 AM.
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  2. #2
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    1) Do you have to do this by definition, or have you already seen the comparison test?

    2) If you want to do this by definition, then you didn't prove anything. You stated that \displaystyle S_m = \sum_{n=1}^{m} \frac{2}{n^2 +n} is convergent, which means \displaystyle \lim_{m \to \infty} S_m = \sum_{n=1}^{m} \frac{2}{n^2+n} converges, but that is what you have to prove! What you said is "this series converges because it converges" - this is not a proof.
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    1) Do you have to do this by definition, or have you already seen the comparison test?

    2) If you want to do this by definition, then you didn't prove anything. You stated that \displaystyle S_m = \sum_{n=1}^{m} \frac{2}{n^2 +n} is convergent, which means \displaystyle \lim_{m \to \infty} S_m = \sum_{n=1}^{m} \frac{2}{n^2+n} converges, but that is what you have to prove! What you said is "this series converges because it converges" - this is not a proof.
    Well, I've stated that S_m is convergent since m is finite and fixed. The rest of the "proof" was intended to show that this extends to the case in which n goes to infinity.

    I did try my hand at the comparison test, but wasn't very successful there either. I'll give it another shot and see what I come up with.
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  4. #4
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    If m is finite and fixed then S_m is just a number. It does not depend on n.

    The easiest way to prove this is by using the comparison test. I assume you have proved that \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} converges. Now,

    \frac{2}{n^2+n} \le \frac{2}{n^2} \ \forall n \in \mathbb{N}

    and so

    \displaystyle \sum_{n=1}^{\infty} \frac{2}{n^2 + n} converges iff \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} converges, since they both have positive terms.
    Last edited by Defunkt; August 23rd 2010 at 05:29 AM. Reason: Indices
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  5. #5
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    Wow. I literally just made the connection with that particular series before coming back to see your post.
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