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Thread: [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 $\displaystyle a_n=\frac{n+(-1)^n}{n(n+1)}$.

    Determine whether $\displaystyle \sum a_n$ converges or diverges.

    My solution:

    Define

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

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

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

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

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

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

    Dividing by $\displaystyle n(n+1)$, we have

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

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

    $\displaystyle b_n<a_n$

    when $\displaystyle n$ is odd.

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

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

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

    $\displaystyle b_n<a_n$

    when $\displaystyle n$ is even.

    Thus, whether $\displaystyle n$ is odd or even, we have $\displaystyle 0<b_n<a_n$, with $\displaystyle \sum b_n$ divergent. By the comparison test, therefore $\displaystyle \sum a_n$ 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 $\displaystyle a_n=\frac{n+(-1)^n}{n(n+1)}$.
    Notice that $\displaystyle a_n = \frac{n+(-1)^n}{n(n+1)} \geq \frac{n-1}{n(n+1)} \geq \frac{n - \tfrac{1}{2}n}{n(n+1)} = \frac{1}{2(n+1)} \geq \frac{1}{2(n+n)} = \frac{1}{4n}$ and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n} = \text{me}$.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Notice that $\displaystyle a_n = \frac{n+(-1)^n}{n(n+1)} \geq \frac{n-1}{n(n+1)} \geq \frac{n - \tfrac{1}{2}n}{n(n+1)} = \frac{1}{2(n+1)} \geq \frac{1}{2(n+n)} = \frac{1}{4n}$ and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n} = \text{me}$.
    So then you agree that this sequence diverges, yes?

    Have I somehow misunderstood the comparison test? If $\displaystyle \sum b_n=\infty$ and $\displaystyle 0<b_n<a_n$ for large $\displaystyle n$, then $\displaystyle \sum a_n$ is divergent, right?
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  4. #4
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    Also note that $\displaystyle \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 $\displaystyle \sum a_n$ 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 $\displaystyle \sum a_n$ 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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