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Math Help - Ratio and Root Tests

  1. #1
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    Ratio and Root Tests

    Good morning All,

    I am working on ratio and root tests for infinite series. I would like to post what I have done, and see if anyone agrees with it. Please see below:


     <br />
\displaystyle\sum_{n=2}^{\infty}(-1)^n\frac{123^n}{n!}<br />

    By the Ratio Test...

     <br />
\displaystyle\lim_{n\rightarrow\infty}\mid\frac{(-1)^{(n+1)}123^{(n+1)}}{(n+1)!}\frac{n!}{(-1)^{n}123^{n}}\mid<br />

    And...

     <br />
\displaystyle\lim_{n\rightarrow\infty}\frac{123n!}  {(n+1)!} = \lim_{n\rightarrow\infty}\frac{123n!}{(n+1)n!} = \lim_{n\rightarrow\infty}\frac{123}{n+1} = 0<br />

    Therefore...

    The orignal series \displaystyle\sum_{n=2}^{\infty}(-)^n\frac{123^n}{n!} is Absolutely Convergent.

    Are we good here?



    My second problem is:

    \displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})

    I'm not really sure where to start here. Would I apply the root test to this problem?
    Last edited by MechEng; August 4th 2010 at 09:42 AM.
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  2. #2
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    The first problem looks good to me, and you have the right idea for the second. What do you get when you try to take the n-th root?
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  3. #3
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    Thanks!

    When I try to take the nth root i get the following:

    \displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})


    \displaystyle\lim_{n\rightarrow\infty}|[(-1)^n(\sqrt{n+1}-\sqrt{n})]^{\frac{1}{n}}|


    =\displaystyle\lim_{n\rightarrow\infty}|(-1)^{\frac{n}{n}}((n+1)^{\frac{1}{2n}}-(n)^{\frac{1}{2n}})|


    =\displaystyle|(-1)^1((n+1)^0-(n)^0)|


    =\displaystyle|(-1)((1)-(1))|


    =\displaystyle|-1+1|


    =\displaystyle|0|


    =\displaystyle0


    That can't be right??? I have three problems; one is supposed to be divergent, one conditionally convergent, and one absolutely convergent. Since my first problem was absolutely convergent (and I was able to follow the entire process for the first problem) I would have thought that the second problem would be either conditonally convergent or divergent.

    ...so, where did I go wrong?
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  4. #4
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    Have you studied the alternating series test? If so that is what is to be used.
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  5. #5
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    I have reviewed the alternating series test.

    I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem.

    Is this reasonable?
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  6. #6
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    Quote Originally Posted by MechEng View Post
    I have reviewed the alternating series test.

    I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem.
    Is this reasonable?
    Did you notice that \displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?
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  7. #7
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    Quote Originally Posted by Plato View Post
    Did you notice that \displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?
    Shoot... I did not notice that. Let me try this again...
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  8. #8
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    When I try to take the nth root i get the following:

    \displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})   =   \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}


    =\displaystyle\lim_{n\rightarrow\infty}|[\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}]^{\frac{1}{n}}|


    =\displaystyle\lim_{n\rightarrow\infty}|\frac{(-1)^\frac{n}{n}}{(n+1)^\frac{1}{2n}+(n)^\frac{1}{2n  }}}|


    =\displaystyle|\frac{(-1)^1}{(n+1)^0+(n)^0}|


    =\displaystyle|\frac{-1}{1+1}|   =   \frac{1}{2}


    Am I losing it again somewhere?

    Or, are you implying that I use the Ratio Test?

    Thanks for the help.
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  9. #9
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    Use the alternating series test.
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  10. #10
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    Ok, I am waiting to hear back with respect to whether or not using the alternating series test is acceptable.

    In the interim, I have worked out the last of three problems and concluded that:


    \displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{(2n)!}{3^n(n!)}

    By Ratio Test...

    \displaystyle\lim_{n\rightarrow\infty}|\frac{(-1)^{(n+1)}(2(n+1))!}{3^{(n+1)}((n+1)!)}\frac{3^n(n  !)}{(-1)^n(2n)!}|

    =\displaystyle\lim_{n\rightarrow\infty}\frac{4n^2+  6n+2}{3n+3}   =   \infty

    Therefore...

    By the Ratio test, the series Diverges




    So, this leaves me with my remaining series that is conditionally convergent. And, this means that if I am able to use the Ratio or Root Test, I should wind up with a limit equal to one... right?
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  11. #11
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    Plato,

    I am noticing a trend with the lesson plans that I am following; the questions seem to be quite difficult using the methods covered in that section, but following section will introduce a method that makes the problem quite simple to do. I assume this one is the same.

    I am not a huge fan of the way the text book is organized, but i am nearly done with it so it's not worth complaining now.

    I appreciate your patience. I used to tutor physics in college, so I know it can be trying when someone keeps mising the fundamental ideas. Thank you for continuing to work with me.
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  12. #12
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    All tests are now available to use on this problem, so I will heed you advice, Plato, and try the alternating series test...
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  13. #13
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    You should find that it in conditional but not absolute.
    To do the later, use the limit comparison test using \dfrac{1}{\sqrt{n}}
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  14. #14
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    Easy question I hope, but...

    Since one of the requirements for the limit comparison test is that the series are both positive, does that mean that my original series becomes...

    \displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1  }+\sqrt{n}}
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  15. #15
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    Please correct me if I am out of whack here...

    Let \displaystyle\sum a_n  =  \frac{1}{\sqrt{n+1}+\sqrt{n}}

    So...

    \displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq  rt{n+1}+\sqrt{n}}=0

    By Alternating Series Test... Convergent

    Let \displaystyle\sum b_n  =  \frac{1}{\sqrt{n}} A Divergent P-series

    So...

    \displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq  rt{n+1}+\sqrt{n}}\frac{\sqrt{n}}{1}

    =\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\s  qrt{n+1}}=0

    By Limit Comparison Test... Divergent

    This doesn't make sense, does it?

    Would it make ore sense if the order of the tests was reversed?
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