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Homework Help: Using the Ratio Test to see if a series converges or diverges?

  1. Nov 22, 2012 #1
    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) [itex]\sum^{∞}_{n=1}\frac{3^{n}}{n^{n}}[/itex]

    (b) [itex]\sum^{∞}_{n=1}\frac{n!}{n^{\frac{n}{2}}}[/itex]


    2. Relevant equations

    [itex]if\:lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}}) < 1 \Rightarrow\sum^{∞}_{n=1}\:-\:converges[/itex]

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

    [itex]0\leq \sum^{∞}_{n=1}(a_n)\leq \sum^{∞}_{n=1}(b_n)[/itex]
    [itex]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\vdots[/itex]
    [itex]if\:\sum^{∞}_{n=1}(a_n)\:-\:diverges\:\Rightarrow\:\sum^{∞}_{n=1}(b_n)\:-\:diverges[/itex]

    [itex]if\:\sum^{∞}_{n=1}(b_n)\:-\:converges\:\Rightarrow\:\sum^{∞}_{n=1}(a_n)\:-\:converges[/itex]

    3. The attempt at a solution

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

    I ended up with:
    [itex]lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}})\:=\:3lim_{n\rightarrow∞}(\frac{n^{n}}{(n+1)(n+1)^{n}})[/itex]

    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 [itex]a_{n}[/itex] = sequence of partial sums then plugged everything into ratio test formula.

    I ended up with:
    [itex]lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}})\:=\:lim_{n\rightarrow∞}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\sqrt{n+1}})\:+\:lim_{n\rightarrow∞}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}{2}}\sqrt{n+1}})[/itex]

    I know that:

    [itex]lim_{n\rightarrow∞}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\sqrt{n+1}})\:=\:∞[/itex]

    And that:

    [itex]lim_{n\rightarrow∞}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}{2}}\sqrt{n+1}})\:=\:0[/itex]

    Hence the overall limit = ∞ 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.
     
  2. jcsd
  3. Nov 22, 2012 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    The ratio test doesn't use the sequence of partial sums. It just compares ##a_{n+1}## with ##a_n##. Your ratio of$$
    \frac{3n^n}{(n+1)(n+1)^{n}}$$ is the right calculation. What happens if you overestimate the ##n^n## in the numerator with ##(n+1)^{n}##?
    For (b) I get a ratio of$$
    (n+1)^{\frac 1 2}\left(\frac n {n+1}\right)^{\frac n 2}$$The right half of that has something to do with the number ##e##. Is that enough of a hint?
     
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