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Thread: Infinite Series Problem

  1. #1
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    Infinite Series Problem

    I'm sure this problem is obvious, but I don't know where to start.

    Determine whether the series converges or diverges and give a good explanation how you arrived at your conclusion.

    $\displaystyle \Sigma^{\infty}_{n = 1}n\sin{\frac{1}{n}}$

    Also, for $\displaystyle \Sigma^{\infty}_{n = 1}\frac{1}{3 + n^2}$, I know I can use the integral test to show it converges, but can I also use the comparison test, comparing it to $\displaystyle \Sigma^{\infty}_{n = 1}\frac{1}{n^2}$?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by mturner07 View Post
    $\displaystyle \Sigma^{\infty}_{n = 1}n\sin{\frac{1}{n}}$
    the limit of the general term doesn't go to zero, so it diverges.

    Quote Originally Posted by mturner07 View Post
    Also, for $\displaystyle \Sigma^{\infty}_{n = 1}\frac{1}{3 + n^2}$, I know I can use the integral test to show it converges, but can I also use the comparison test, comparing it to $\displaystyle \Sigma^{\infty}_{n = 1}\frac{1}{n^2}$?
    yes.
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  3. #3
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    Isn't $\displaystyle \lim_{n \to \infty}n\sin{\frac{1}{n}} = 0$, since $\displaystyle \infty *\sin{0} = 0$?
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  4. #4
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    Quote Originally Posted by mturner07 View Post
    Isn't $\displaystyle \lim_{n \to \infty}n\sin{\frac{1}{n}} = 0$, since $\displaystyle \infty *\sin{0} = 0$?
    $\displaystyle (\infty \cdot 0)$ is an indeterminate form.

    $\displaystyle \lim_{n \to \infty} n \cdot \sin\left(\frac{1}{n}\right) = \lim_{\frac{1}{n} \to 0} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} = 1$
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  5. #5
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    D'oh, thanks.
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