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Math Help - [SOLVED] Is my textbook wrong? Series convergence vs divergence.

  1. #1
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    [SOLVED] Is my textbook wrong? Series convergence vs divergence.

    The problem:

    Define a_n=\frac{n+(-1)^n}{n(n+1)}.

    Determine whether \sum a_n converges or diverges.

    My solution:

    Define

    b_n=\frac{1}{2n}.

    We will use the comparison test to show that since \sum b_n diverges, so too does \sum a_n diverge.

    Observe that if 3<n, we may divide both sides by 2 to yield

    \frac{3}{2}<\frac{n}{2}.

    Adding \frac{n}{2}-1 to both sides gives us

    \frac{n+1}{2}=\frac{3}{2}+\frac{n}{2}-1<\frac{n}{2}+\frac{n}{2}-1=n-1.

    Dividing by n(n+1), we have

    \frac{1}{2n}<\frac{n-1}{n(n+1)}.

    Since b_n=\frac{1}{2n} and a_n=\frac{n-1}{n(n+1)} when n is odd, then we have

    b_n<a_n

    when n is odd.

    Now observe that since \frac{1}{2}<1, then dividing by n gives us

    \frac{1}{2n}<\frac{1}{n}=\frac{n+1}{n(n+1)}.

    Since b_n=\frac{1}{2n} and a_n=\frac{n+1}{n(n+1)} when n is even, then we have

    b_n<a_n

    when n is even.

    Thus, whether n is odd or even, we have 0<b_n<a_n, with \sum b_n divergent. By the comparison test, therefore \sum a_n is likewise divergent.

    The textbook tells me I am wrong, and that this series is actually convergent.

    Any ideas what went wrong, here?

    Thanks!
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    The problem:
    Define a_n=\frac{n+(-1)^n}{n(n+1)}.
    Notice that a_n = \frac{n+(-1)^n}{n(n+1)} \geq \frac{n-1}{n(n+1)} \geq \frac{n - \tfrac{1}{2}n}{n(n+1)} = \frac{1}{2(n+1)} \geq \frac{1}{2(n+n)} = \frac{1}{4n} and \sum_{n=1}^{\infty}\frac{1}{4n} = \text{me}.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Notice that a_n = \frac{n+(-1)^n}{n(n+1)} \geq \frac{n-1}{n(n+1)} \geq \frac{n - \tfrac{1}{2}n}{n(n+1)} = \frac{1}{2(n+1)} \geq \frac{1}{2(n+n)} = \frac{1}{4n} and \sum_{n=1}^{\infty}\frac{1}{4n} = \text{me}.
    So then you agree that this sequence diverges, yes?

    Have I somehow misunderstood the comparison test? If \sum b_n=\infty and 0<b_n<a_n for large n, then \sum a_n is divergent, right?
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  4. #4
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    Also note that  \sum \frac{n + (-1)^n}{n(n+1)} = \sum \frac{n}{n(n+1)} + \sum \frac{(-1)^n}{n(n+1)} . The first sum diverges while the second one converges, so the original sum must diverge as well.
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  5. #5
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    Yes, I think that everyone would agree that \sum a_n diverges. Your book is mistaken, or maybe you are looking at the wrong answer. Do they provide a "proof" for their claim?
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  6. #6
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    Quote Originally Posted by putnam120 View Post
    Yes, I think that everyone would agree that \sum a_n diverges. Your book is mistaken, or maybe you are looking at the wrong answer. Do they provide a "proof" for their claim?
    Checked and double-checked. I'm definitely looking at the right answer.

    They do not provide any reasons, or a proof.
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