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Math Help - can someone verify if i'm right about these series

  1. #1
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    can someone verify if i'm right about these series

    I have to find whether these series converge.

    \sum {\frac {n+1} {n^2 + 1}} by using the limit comparison test, and using \sum {b_n} = \frac {1} {n} i found that it diverges.

    As for this example: \sum {\frac {n!} {n^n}} i used the ratio test and found it diverges also.

    Can someone check if i am correct, and also help me with this one
    \sum {\frac {1} {2^n - 1 + cos^2(n^3)}}
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  2. #2
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    Quote Originally Posted by Cato View Post
    I have to find whether these series converge.

    \sum {\frac {n+1} {n^2 + 1}} by using the limit comparison test, and using \sum {b_n} = \frac {1} {n} i found that it diverges.
    This diverges so you are correct.

    As for this example: \sum {\frac {n!} {n^n}} i used the ratio test and found it diverges also.
    This is wrong. The ratio that you should get is \left( \frac{n}{n+1} \right)^n. The limit of this is 1/e < 1. Therefore it converges.

    Can someone check if i am correct, and also help me with this one
    \sum {\frac {1} {2^n - 1 + cos^2(n^3)}}
    \frac{1}{2^n - 1 + \cos^2(n^3)} \leq \frac{1}{2^n - 1}
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  3. #3
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    Thank you, so \frac{1}{2^n - 1} would then diverge when n tends to infinity?
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  4. #4
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    Quote Originally Posted by Cato View Post
    I have to find whether these series converge.

    \sum {\frac {n+1} {n^2 + 1}} by using the limit comparison test, and using \sum {b_n} = \frac {1} {n} i found that it diverges.

    As for this example: \sum {\frac {n!} {n^n}} i used the ratio test and found it diverges also.

    Can someone check if i am correct, and also help me with this one
    \sum {\frac {1} {2^n - 1 + cos^2(n^3)}}
    You got it right for the first.
    For the last one, use the fact that \cos x\leq 1. If i give you a little bit more hint it will be obvious. So try first and let see if you get it or not


    Edit: sorry I don't know if someone already give a reply (the ad always deceives me as if there is no reply to the post).
    I think it is also bound by 1/2^n which is converges (geometric series).
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