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Math Help - Series problems--conv. or div?

  1. #1
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    Series problems--conv. or div?

    I have four calculus problems that I haven't been able to figure out. The general question for them is "Test the series for convergence or divergence by explicitly showing the test being used."
    I've tried the Limit comparison test, Integral test, Ratio test, and Root test. All failed for me. Here's the questions: (all are series from n=1 to infinity, i didn't know how to use this forum's math type)

    1. 10^n/n!

    2. sin(n)

    3. (2n)^n/n^(2n)

    4. n^2/e^((n)^3)


    Any and all help is greatly appreciated!
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  2. #2
    Super Member wingless's Avatar
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    See here for using Latex typesetting.


    Quote Originally Posted by KennyYang View Post
    1. 10^n/n!
    Use Cauchy Condensation Test. I couldn't find a better way to do it, but there must be..
    (Notice that the series is decreasing after n=10)

    Quote Originally Posted by KennyYang View Post
    2. sin(n)
    This sequence diverges, so the series diverge too.
    Last edited by wingless; March 8th 2008 at 02:07 PM.
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  3. #3
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    Which test did you use for sin(n)? It's clear that it's divergent since it just oscillates, but which test will mathematically show this? My professor is very particular about how his students show work.
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by KennyYang View Post
    Which test did you use for sin(n)? It's clear that it's divergent since it just oscillates, but which test will mathematically show this? My professor is very particular about how his students show work.
    It's called Divergence Test or Term Test.

    If \lim \left ( a_n \right) is not zero, then \sum^{\infty}_{n=1} a_n diverges.
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  5. #5
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    Quote Originally Posted by KennyYang View Post
    Which test did you use for sin(n)? It's clear that it's divergent since it just oscillates, but which test will mathematically show this? My professor is very particular about how his students show work.
    I hope that your professor is most concerned that you know and understand the basic ideas here. It seems that you have a great deal of study ahead of you if you are to meet that goal. Thus, if I were you I would not be so concerned with presentation, but rather exhibiting understanding of how series work.
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  6. #6
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    Quote Originally Posted by KennyYang View Post
    I have four calculus problems that I haven't been able to figure out. The general question for them is "Test the series for convergence or divergence by explicitly showing the test being used."
    I've tried the Limit comparison test, Integral test, Ratio test, and Root test. All failed for me. Here's the questions: (all are series from n=1 to infinity, i didn't know how to use this forum's math type)

    1. 10^n/n!

    2. sin(n)

    3. (2n)^n/n^(2n)

    4. n^2/e^((n)^3)



    Any and all help is greatly appreciated!
    As you can see first example for small n is growing but for higher n factorial faster almost than any useful function except Bessel asimptotic function.
    If you have used D'Alambert:

    <br />
|\frac{10}{n+1}|< 1 \;when\;\; n \longrightarrow \infty\;;\;\;\;<br />
  \sum_{n=1}^{\infty} \frac{10^{n}}{n!}=e^{10} - 1<br />
    Also you can eveluate with Stirling's formula:
    <br />
n!=\sqrt{2\pi n}\; \frac{n^{n}}{e^{n}} (1+\frac{1}{12 n}+\frac{1}{288n^2}+...)<br />
    and than you can do anything you want

    <br />
\sum^{\infty}_{n=1} Sin(n)=\sum^{\infty}_{n=1} \sum^{\infty}_{i=1} \frac{(-1)^{2i+1} n^{2i+1}}{(2i+1)!}<br />
    That sum does not converge; converge only for |n| < 1.

    For third example:
    Use Cauchy's (roots test) and at the and you got
    lim(n \rightarrow \infty)\; \frac{2}{n} = 0
    and goes to 0 when n rise to infinity.

    And the last one:
    Use comparison test and rough estimation (it is only estimation)
    <br />
\frac{n^2}{e^{n^3}} < \frac{1}{n^{n}}<br />
    Why you can use this estimation?
    Becuse you know that
    <br />
\frac{1}{n^{n}} < \frac{1}{2^{n}}<br />

    and also

    <br />
\sum_{n=0}^{\infty} \frac{1}{2^{n}} = 2<br />

    I try my best to make clear and
    Good luck with professor.
    Miloš
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