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Math Help - Converge of Diverge

  1. #1
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    Converge of Diverge

    Determine whether those series are converge and diverge?if possible, find the limit of the convergent series.
    1. (n^+2^n)/(n+3^n) from n=1 to infinite
    2. log(8n^2 + 1) + 2 log(1/n)
    3. [√(n^4+ 3n^2+ 1)] - n^2 -1 ( just this sequence, not "sum")

    Last edited by Jojo123; November 15th 2008 at 04:55 PM.
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  2. #2
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    Hello, Jojo123!

    2)\;\;S \;=\;\sum^{\infty}_{n=1}\bigg[\ln(8n^2 + 1) + 2\log\left(\tfrac{1}{n}\right)\bigg]

    We have: / \ln(8n^2+1) + \ln\left(\tfrac{1}{n}\right)^2 \;=\;\ln(8n^2+1) + \ln\left(\tfrac{1}{n^2}\right) \;=\;\ln\bigg[(8n^2+1)\cdot\tfrac{1}{n^2}\bigg]

    . . . . . . = \;\ln\left(\frac{8n^2+1}{n^2}\right) \;=\;\ln\left(\frac{8n^2}{n^2} + \frac{1}{n^2}\right) \;=\;\ln\left(8 + \frac{1}{n^2}\right)


    Hence: . S \;=\;\sum^{\infty}_{n=1}\ln\left(8 + \tfrac{1}{n^2}\right)

    . . . . . . S \;=\;\ln\left(8 + \tfrac{1}{1^2}\right) + \ln\left(8 + \tfrac{1}{2^2}\right) + \ln\left(8 + \tfrac{1}{3^2}\right) + \hdots

    . . . . . . S \;=\;\ln\underbrace{\bigg[\left(8 + \tfrac{1}{1}\right)\left(8 + \tfrac{1}{4}\right)\left(8 + \tfrac{1}{9}\right) \hdots \bigg]}_{\text{infinite product}}

    This product has an infinite number of factors which are greater than 8.
    . . Hence, the product diverges.

    Therefore, the series diverges.

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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jojo123 View Post
    Determine whether those series are converge and diverge?if possible, find the limit of the convergent series.
    1. (n2+2n)/(n+3n) from n=1 to infinite
    2. log(8n2 + 1) + 2 log(1/n)
    3. [√(n4+ 3n2+ 1)] - n2 -1

    For the other two consider if the kernel of the summation is a null sequence or not.
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  4. #4
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    For the other two consider if the kernel of the summation is a null sequence or not.
    What do you mean ?
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    What do you mean ?
    Yes, thank you very much Moo. I need to learn to say things in a more understandable fashion when helping someone online. To original poster, what I said just means that make sure that sequence being summed tends to zero. Otherwise it diverges.
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