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Ratio and root tests

  1. Aug 20, 2009 #1
    [tex]
    1. \sum_{n=1}^{\infty}ntan\frac{1}{n}
    [/tex]
    dont know where to start here
    [tex]
    2. \frac{1*3*5*****(2n-1)}{n!}[/tex]
    [tex]
    \frac{2n-1}{n!}[/tex]
    [tex]
    \frac{2(n+1)-1}{(n+1)!}*\frac{n!}{2n-1}[/tex]
    [tex]
    \frac{2n+1}{(n+1)(2n-1)}[/tex]
    [tex]->0[/tex]
    my book is showing divergence
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 20, 2009 #2

    Gib Z

    User Avatar
    Homework Helper

    It doesn't appear you applied the tests correctly?

    The ratio test says that the sum [itex]\sum_{n=0}^{\infty} a_n[/itex] converges if [itex]\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right|[/itex] is less than 1, diverges if it is greater than one, and is inconclusive if it equals exactly 1.

    The root test says the sum will converge depending on the result of [itex]\lim_{n\to\infty} (a_n)^{1/n}[/itex] with similar conditions as above.

    From what I can see, you didn't apply either. Your book is correct. An easier way to do this is using the fact that the terms must approach zero to converge.
     
  4. Aug 20, 2009 #3
    (1) This series diverges. Show that

    [tex]\lim_{n \rightarrow \infty}\frac{\tan (1/n)}{(1/n)}[/tex]​

    is not 0.

    (2) I assume you mean the series

    [tex]\sum_{n=1}^{\infty}\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}[/tex]​

    Any series that involves an iterative product (like this one), the ratio test is a good first approach. Note factorials are iterative products.

    [tex]\frac{a_{n+1}}{a_n}=\frac{1 \cdot 3 \cdot 5 \cdots (2n-1) \cdot (2n + 1)}{(n+1)!} \cdot \frac{n!}{1 \cdot 3 \cdot 5 \cdots (2n-1)}[/tex]​

    Hopefully it is smooth sailing from here.

    --Elucidus
     
  5. Aug 21, 2009 #4
    [tex]\sum_{n=1}^{\infty}\frac{1\cdot3\cdot5\cdot
    \cdot\cdot(2n-1)}{n!}[/tex]
    this is the first problem I have seem with listed factors
    I applied the ratio test the same as if to
    [tex]\sum_{n=1}^{\infty}\frac{2n-1}{n!}[/tex]
    it seems as if the test has been applied diffrently here forgive me if I am missing somthing obvious but
    [tex]a_{n+1}=\frac{1\cdot3\cdot5\cdot
    \cdot\cdot(2n-1)(2n+1)}{(n+1)!}[/tex]
    are we simply proceeding into the next iteration by listing (2n-1) along with (2n+1)
    [tex]\frac{a_{n+1}}{a_{n}}=\frac{1\cdot3\cdot5\cdot
    \cdot\cdot(2n-1)(2n+1)}{(n+1)!}\cdot\frac{n!}{1\cdot3\cdot5\cdot
    \cdot\cdot(2n-1)}=\frac{2n+1}{n+1}=2[/tex]
    to show the series is divergent
    for the trig series i see the the limit tending to 1 is this a conditon that must be satisfied
     
    Last edited: Aug 21, 2009
  6. Aug 21, 2009 #5

    zcd

    User Avatar

    The condition for the ratio test is that [tex]|\lim_{n\to \infty}\frac{a_{n+1}}{a_{n}}|<1[/tex] for convergence, >1 for divergence, and =1 for indeterminant. Since the limit converges to a value of 2>1, the series is divergent by ratio test.
     
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