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Math Help - Can someone help me with these series problems?

  1. #1
    s3a
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    Can someone help me with these series problems?

    How do I show that diverges? Also, how do I show that converges? If you're too lazy for a full response, I think that just naming a test would be enough for such a problem.

    Any input would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    How do I show that diverges? Also, how do I show that converges? If you're too lazy for a full response, I think that just naming a test would be enough for such a problem.

    Any input would be greatly appreciated!
    Thanks in advance!
    For the first one, you can use comparison with an integral (note that \int\frac{dx}{x\ln x}=\ln(\ln x)). (or Cauchy condensation test)

    For the second one, convergence is very quick, so any basic test will work.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by s3a View Post
    How do I show that diverges? Also, how do I show that converges? If you're too lazy for a full response, I think that just naming a test would be enough for such a problem.

    Any input would be greatly appreciated!
    Thanks in advance!
    Use the integral test for both


    \int\limits_2^\infty\frac{8\,dx}{3x\ln x}=\frac{8}{3}\lim\limits_{b\to\infty}\int\limits_  2^b\frac{d(\ln x)}{\ln x}=\left.{\frac{8}{3}\lim\limits_{b\to\infty}\ln\l  n x}\right|_2^b = \frac{8}{3}\lim\limits_{b \to\infty}\left(\ln\ln b - \ln \ln 2\right) = \infty.


    \int\limits_1^\infty {xe^{-x^2}\,dx}=-\frac{1}{2}\mathop {\lim }\limits_{b\to\infty} \int\limits_1^b{e^{-x^2}\,d\left(-x^2\right)}= \left. { - \frac{1}{2}\lim\limits_{b\to\infty}e^{-x^2}}\right|_1^b = -\frac{1}{2}\lim\limits_{b\to\infty}\left(e^{-b^2}-e^{-1}\right) = \frac{e^{-1}}{2}.
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  4. #4
    s3a
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    When I do the first one, I did it with a u substitution u = ln(n)
    and my last step is 8/3 * lim t->inf ln(t) - ln(0) but I said ln(0) not ln(0.00000000000000000001...) which does not exist. So, what does that mean or did I do something wrong?
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by s3a View Post
    When I do the first one, I did it with a u substitution u = ln(n)
    and my last step is 8/3 * lim t->inf ln(t) - ln(0) but I said ln(0) not ln(0.00000000000000000001...) which does not exist. So, what does that mean or did I do something wrong?
    when you use a u-substution, you have:

    \int \frac{8}{3n.ln(n)} dx

    let u = ln(n) \implies du = \frac{1}{n} dn \implies dn = n.du

    so,

    \frac{8}{3} \int \frac{n}{n. u} du = \frac{8}{3} \int{1}{u} du = \frac{8}{3} log(u) = \frac{8}{3} ln(ln(n))

    you can use the limits given to see that the integral diverges.. this has been done in the above post by DeMath
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by s3a View Post
    When I do the first one, I did it with a u substitution u = ln(n)
    and my last step is 8/3 * lim t->inf ln(t) - ln(0) but I said ln(0) not ln(0.00000000000000000001...) which does not exist. So, what does that mean or did I do something wrong?
    For substitution u=\ln{x} :

    If x=2 then u=\ln{2}

    If x\to\infty then also u\to\infty

    I.e. you have

    \frac{8}{3}\int\limits_2^\infty\frac{dx}{x\ln{x}}=  \left\{\begin{gathered}u=\ln{x}\hfill\\du=\frac{dx  }{x}\hfill\\\end{gathered}\right\}=\frac{8}{3}\int  \limits_{\ln2}^\infty\frac{du}{u}
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  7. #7
    s3a
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    Oh, my mistake was just that I thought the interval was from [1,inf] instead of [2,inf].
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