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Thread: Ratio test question

  1. #1
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    Ratio test question

    hi everyone

    im stuck with this question, need help to clarify..

    Determine this series is absolutely convergent or diverges by using ratio test.

    \sum ^\infty_{n=1} \frac{(-1)^n e^\frac{1}{n}}{n^3}

    i manage to get this
    \frac{\frac{(-1)^{n+1} e^\frac{1}{n+1}}{(n+1)^3}}{\frac{(-1)^n e^\frac{1}{n}}{n^3}}

    \frac{-(n^3)e^\frac{1}{n+1}}{e^\frac{1}{n+1}(n+1)^3}

    how do i complete this working, out of ideas.need help.

    really appreciate all your help & guidance, thank you in advance for all your help & support.
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  2. #2
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    Quote Originally Posted by anderson View Post
    hi everyone

    im stuck with this question, need help to clarify..

    Determine this series is absolutely convergent or diverges by using ratio test.

    \sum ^\infty_{n=1} \frac{(-1)^n e^\frac{1}{n}}{n^3}

    i manage to get this
    \frac{\frac{(-1)^{n+1} e^\frac{1}{n+1}}{(n+1)^3}}{\frac{(-1)^n e^\frac{1}{n}}{n^3}}

    \frac{-(n^3)e^\frac{1}{n+1}}{e^\frac{1}{n+1}(n+1)^3}

    how do i complete this working, out of ideas.need help.

    really appreciate all your help & guidance, thank you in advance for all your help & support.
    You actually need to find \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|

     = \lim_{n \to \infty}\left|\frac{\frac{(-1)^{n + 1}e^{\frac{1}{n + 1}}}{(n + 1)^3}}{\frac{(-1)^n e^\frac{1}{n}}{n^3}}\right|

     = \lim_{n \to \infty}\frac{\frac{e^{\frac{1}{n + 1}}}{(n + 1)^3}}{\frac{e^{\frac{1}{n}}}{n^3}}

     = \lim_{n \to \infty}\frac{n^3e^{\frac{1}{n + 1}}}{(n + 1)^3e^{\frac{1}{n}}}

     = \lim_{n \to \infty}\frac{n^3e^{\frac{1}{n + 1} - \frac{1}{n}}}{(n + 1)^3}

     = \lim_{n \to \infty}\frac{n^3e^{-\frac{1}{n(n + 1)}}}{(n + 1)^3}

     = \lim_{n \to \infty}\frac{n^3}{(n + 1)^3e^{\frac{1}{n(n + 1)}}}


    Now yo might want to try using L'Hospital's Rule to simplify...
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  3. #3
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    thank you for guiding. got no idea how to use L Hospital rule to simplify this question, can some one help me.

    thank you for all your help & guidance.
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  4. #4
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    There is no need to use L'Hopital's rule. Separate it into two parts:
    <br />
= \frac{n^3}{(n + 1)^3}e^{-\frac{1}{n(n + 1)}}<br />

    Now, \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^3 should be easy. As n goes to infinity, \frac{1}{n(n+1)} goes to 0. I think you have a problem applying the ratio test here!
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  5. #5
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    thank you for replying.
    <br /> <br />
\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^3<br />
= 1

    is that correct? then how do i simplify the other limit , e^{-\frac{1}{n(n + 1)}}

    need some guide & help, thank you in advance for all help & support.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    There is no need to use L'Hopital's rule. Separate it into two parts:
    <br />
= \frac{n^3}{(n + 1)^3}e^{-\frac{1}{n(n + 1)}}<br />

    Now, \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^3 should be easy. As n goes to infinity, \frac{1}{n(n+1)} goes to 0. I think you have a problem applying the ratio test here!
    Wouldn't that mean that e^{-\frac{1}{n(n + 1)}} tends to e^0 = 1?

    So really, all you have to worry about is the \left(\frac{n}{n + 1}\right)^3 = \left(1 - \frac{1}{n + 1}\right)^3 which tends to 1.

    Ohhh, so it all tends to 1 - so I agree with you that the ratio test fails...


    Why not try LAST instead?
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  7. #7
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    did you mean

    e^-\frac{1}{n^2+n}
    e^-\frac{\frac{1}{n^2}}{\frac{n^2}{n^2}+\frac{n}{n^2}  }
    e^0 = 1 ?

    thank you for confirming, sorry to trouble everyone...

    thank you in advance for all your help.
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  8. #8
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    Quote Originally Posted by anderson View Post
    did you mean

    e^-\frac{1}{n^2+n}
    e^-\frac{\frac{1}{n^2}}{\frac{n^2}{n^2}+\frac{n}{n^2}  }
    e^0 = 1 ?

    thank you for confirming, sorry to trouble everyone...

    thank you in advance for all your help.
    Surely if -\frac{1}{n(n + 1)} tends to 0 then e^{-\frac{1}{n(n + 1)}} tends to e^0 = 1...
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