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Math Help - Using the Ratio Test to see if a series converges or diverges?

  1. #1
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    Using the Ratio Test to see if a series converges or diverges?

    1. The problem statement, all variables and given/known data


    Use the Ratio Test for series to determine whether each of the following series converge or diverge. Make Reasoning Clear.


    (a) \sum^{infinity}_{n=1}\frac{3^{n}}{n^{n}}


    (b) \sum^{infinity}_{n=1}\frac{n!}{n^{\frac{n}{2}}}




    2. Relevant equations


    if\:lim_{n\rightarrow infinity}(\frac{a_{n+1}}{a_{n}}) < 1 \Rightarrow\sum^{infinity}_{n=1}\:-\:converges


    if\:lim_{n\rightarrow infinity}(\frac{a_{n+1}}{a_{n}})\in[0,1)\Rightarrow \sum^{infinity}_{n=1}\:-\:diverges


    0\leq \sum^{infinity}_{n=1}(a_n)\leq \sum^{infinity}_{n=1}(b_n)
    \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:  \:\:\:\:\vdots
    if\:\sum^{infinity}_{n=1}(a_n)\:-\:diverges\:\Rightarrow\:\sum^{infinity}_{n=1}(b_n  )\:-\:diverges


    if\:\sum^{infinity}_{n=1}(b_n)\:-\:converges\:\Rightarrow\:\sum^{infinity}_{n=1}(a_  n)\:-\:converges


    3. The attempt at a solution


    (a) I let a_{n} = sequence of partial sums then plugged everything into ratio test formula.


    I ended up with:
    lim_{n\rightarrow infinity}(\frac{a_{n+1}}{a_{n}})\:=\:3lim_{n|:tend  s\:to\:infinity}(\frac{n^{n}}{(n+1)(n+1)^{n}})


    I know that the limit equals 0 hence the series converges but not quite sure how to show that the limit cancels down to show 0 is "obvious"


    Any help would be great, thanks.


    (b) I let a_{n} = sequence of partial sums then plugged everything into ratio test formula.


    I ended up with:
    lim_{n\rightarrow}(\frac{a_{n+1}}{a_{n}})\:=\:lim_  {n\rightarrow infinity}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\s  qrt{n+1}})\:+\:lim_{n\rightarrow infinity}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}  {2}}\sqrt{n+1}})


    I know that:


    lim_{n\rightarrow infinity}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\s  qrt{n+1}})\:=\:infinity


    And that:


    lim_{n\rightarrow infinity}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}  {2}}\sqrt{n+1}})\:=\:0


    Hence the overall limit = infinity and so the series diverges; but, once again i'm not quite sure how to show that how the limits cancels down to show ∞ and 0 is "obvious"


    Once again, any help would be great, thanks.
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  2. #2
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    Re: Using the Ratio Test to see if a series converges or diverges?

    For a) notice that \displaystyle \begin{align*} a_n = \frac{3^n}{n^n} \end{align*} and \displaystyle \begin{align*} a_{n+1} = \frac{3^{n+1}}{(n + 1)^{n+1}} \end{align*}. Then

    \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{\frac{3^{n+1}}{(n+1)^{n+1}}}{\frac{3^n}{n^n}  }  \right| \\ &= \lim_{n \to \infty} \frac{3^{n+1}\, n^n}{3^n\left (n + 1 \right)^{n+1}} \\ &= \lim_{n \to \infty} \left[ \frac{3}{(n+1)} \left( \frac{n}{n+1} \right)^n \right] \\ &= \lim_{n \to \infty} \frac{3}{n+ 1} \cdot \lim_{n \to \infty} \left[ \left( \frac{n}{n+1} \right)^n \right] \\ &= 0 \lim_{n \to \infty} e^{\ln{\left[ \left( \frac{n}{n+1} \right)^n \right]}} \\ &= 0 \, e^{ \lim_{n \to \infty} \frac{\ln{\left( 1 - \frac{1}{n+1} \right)}}{\frac{1}{n}} } \\ &= 0\, e^{\lim_{n \to \infty} \frac{\frac{1}{n} - \frac{1}{n+1}}{-\frac{1}{n^2}}} \textrm{ by L'Hospital's Rule} \\ &= 0 \, e^{\lim_{n \to \infty} \frac{\frac{1}{n(n+1)}}{-\frac{1}{n^2}}} \\ &= 0\, e^{\lim_{n \to \infty} -\frac{n^2}{n^2 + n}} \\ &= 0\, e^{\lim_{n \to \infty} -\frac{1}{1 + \frac{1}{n}}} \\ &= 0 \, e^{-1} \\ &= 0 \end{align*}
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    Re: Using the Ratio Test to see if a series converges or diverges?

    Thanks, that's very good and I understand all of it; however, I haven't been formally taught L'Hospitals Rule yet and so am not allowed to use it for this particular problem.

    Any other ideas?
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    Re: Using the Ratio Test to see if a series converges or diverges?

    Say that it's self-taught then.
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    Re: Using the Ratio Test to see if a series converges or diverges?

    Could I equally get away with saying that since the first part of the limit = 0 then it doesn't matter what the second part = since it will still be zero?
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    Re: Using the Ratio Test to see if a series converges or diverges?

    Quote Originally Posted by lmstaples View Post
    Could I equally get away with saying that since the first part of the limit = 0 then it doesn't matter what the second part = since it will still be zero?
    No, because \displaystyle \begin{align*} 0 \times \infty \end{align*} is an entirely different kettle of fish.

    I expect you would have to make use of \displaystyle \begin{align*} \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e \end{align*} somehow...
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    Re: Using the Ratio Test to see if a series converges or diverges?

    Now that standard limit I have been taught and can use. I was trying to implement it before buy didn't get anywhere - I should have kept with it longer.
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    Re: Using the Ratio Test to see if a series converges or diverges?

    Hello, Imstaples!

    Another approach . . .


    \displaystyle \sum^{\infty}_{n=1}\,\frac{3^n}{n^n}

    \text{Ratio} \;=\;\frac{3^{n+1}}{(n+1)^{n+1}}\cdot\frac{n^n}{3^  n} \;=\; \frac{3^{n+1}}{3^n}\cdot\frac{n^n}{(n+1)^{n+1}} \;=\;3\cdot\frac{n^n}{(n+1)(n+1)^n}

    . . . . . =\;\frac{3}{n+1}\cdot\frac{n^n}{(n+1)^n} \;=\;\frac{3}{n+1}\left(\frac{n}{n+1}\right)^n \;=\;\frac{3}{n+1}\left(\frac{1}{1+\frac{1}{n}} \right)^n

    . . . . . =\;\frac{3}{n+1}\cdot\frac{1}{(1+\frac{1}{n})^n}


    \displaystyle \lim_{n\to\infty} \frac{3}{n+1}\cdot\frac{1}{(1+\frac{1}{n})^n} \;=\; \underbrace{\lim_{n\to\infty}\frac{3}{n+1}}_{\text  {This is 0}} \cdot\frac{1}{\underbrace{\lim_{n\to\infty}\left(1  +\tfrac{1}{n} \right)^n}_{\text{This is }e}} \;=\;0\cdot\frac{1}{e} \;=\;0
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  9. #9
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    Re: Using the Ratio Test to see if a series converges or diverges?

    To the OP, I like very much the way you presented the problem. Let me offer a small tip for your use of \LaTeX:

    i) Use the code \infty to get \infty

    ii) Use the code \lim_{n\to\infty} to get \lim_{n\to\infty}
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