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Ratio test

  1. Mar 5, 2016 #1
    1. The problem statement, all variables and given/known data
    when the test is inconclusive when p = 1? when p=1 , the sum of uk will grow bigger , to infinity , right ?

    2. Relevant equations


    3. The attempt at a solution
    let's say uk = 2 , uk_2 = 2 , so as uk_3 ,l uk_4 ....... the sum of all of them will beocme infinity , right?
     

    Attached Files:

  2. jcsd
  3. Mar 6, 2016 #2

    Samy_A

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    When p=1, the test is inconclusive, meaning that the series may or may not converge.
    Your example indeed diverges.
    What about the series defined by ##v_n=\frac{1}{n²}## (for ##n \in \mathbb N,\ n\neq 0##)?
     
  4. Mar 6, 2016 #3
    it will diverge , right ? why it is said to be inconclusive ?
     
    Last edited: Mar 6, 2016
  5. Mar 6, 2016 #4

    Samy_A

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    Yes, it will converge.

    This shows that ##\rho =1## doesn't tell you anything about convergence or not of the series.
    Your series ##u_n## has a ratio ##\rho =1##, and diverges.
    My series ##v_n## has a ratio ##\rho =1##, and converges.

    That is what is meant with inconclusive: when ##\rho =1##, you have to find another method to determine whether the series converges or diverges.
     
  6. Mar 6, 2016 #5
    sorry , i made a typo in the previous post , i mean it will DIVERGE , let's say we have 1/ (100^2) , we have 200 terms of 1/ (100^2) , we will get sum of them = 0.02 , if there are 300 of them , we will get 0.03 .......
     
  7. Mar 6, 2016 #6

    Samy_A

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    No, it really converges.

    If you take the partial sum up to some number ##N##, you will indeed see that:
    ##\displaystyle \sum_{n=1}^N \frac{1}{n²} \geq \sum_{n=1}^N \frac{1}{N²} =\frac{N}{N²}=\frac{1}{N}##.
    But why would that imply divergence?
    (See @Ray Vickson post about the proof that the series converges.)

    For the harmonic series, you can group the terms in such a way that it follows that the limit of the partial sums is +∞. (See @stevendaryl 's post.)
    You can't do this with this series.
     
  8. Mar 6, 2016 #7
    why the small n is replaced by big N ? they are not the same , right ? the sum of them should be N(1/n^2) , right ?
     
  9. Mar 6, 2016 #8

    Samy_A

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    ##N## represent a fixed, chosen number, like you chose 200 or 300 in post #5.
    ##n## is the summation index.

    ##\displaystyle \sum_{n=1}^N \frac{1}{n²}=\frac{1}{1}+\frac{1}{2²}+\frac{1}{3²}+\dots +\frac{1}{(N-1)²}+\frac{1}{N²}##
    If N=300, then
    ##\displaystyle \sum_{n=1}^N \frac{1}{n²}=\sum_{n=1}^{300} \frac{1}{n²}=\frac{1}{1}+\frac{1}{2²}+\frac{1}{3²}+\dots +\frac{1}{299²}+\frac{1}{300²}##
     
  10. Mar 6, 2016 #9

    Mark44

    Staff: Mentor

    N is a fixed number that is reasonably large. n is the index on the summation, taking on the values 1, 2, 3, ..., up to N.

    Also, "text-speak" such as "u" for "you" isn't allowed on this forum.
     
  11. Mar 6, 2016 #10
    why the n inside the (1/n^2) not a fixed number? since we have p (ratio) =1
     
  12. Mar 6, 2016 #11

    Samy_A

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    Why should it?
    The ##n## is not fixed, it is the summation index in ##\displaystyle \sum_{n=1}^{\infty}u_n##.

    The ration ##\rho## is defined by ##\rho=\displaystyle \lim_{n \rightarrow \infty}\frac{|u_{n+1}|}{|u_n|}##.
    ##\rho=1## doesn't mean that ##\frac{|u_{n+1}|}{|u_n|}=1## for some (or all) ##n##, but that the limit of these ratio's is 1.
     
    Last edited: Mar 6, 2016
  13. Mar 6, 2016 #12

    Mark44

    Staff: Mentor

    n isn't fixed -- the exponent 2 is fixed. p is not the ratio in the Ratio Test -- it's the exponent in ##\frac 1 {n^p}##.

    Please take more time to read these tests more carefully.
     
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