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Can anyone explain to me this infinite series problem?

  1. Mar 23, 2017 #1
    1. The problem statement, all variables and given/known data
    eq0015MP.gif empty.gif

    and in this case we have,

    [PLAIN]http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries_files/eq0016MP.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries_files/empty.gif [Broken]

    2. Relevant equations
    I can not see how they get either of the 3 terms or where the 3/4 term comes from.
    Once given the expression, its easy to see it converges to be 3/4.

    3. The attempt at a solution
    Even if I limit shift to either i=0 or i=1, I can not see this solution. Wolfram Alpha confirms 3/4 is the correct answer. Thanks for your help.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Mar 23, 2017 #2

    SammyS

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    Is this from a course?

    If so, what is the main topic of coverage?

    Try partial fraction decomposition of ## \displaystyle \ \frac{1}{i^2-1} \ ##
     
    Last edited by a moderator: May 8, 2017
  4. Mar 23, 2017 #3
    Infinite Series convergence and divergence.

    PFD will give A / i +1 + B / i - 1 and that does not look in any way like the given solution.
     
  5. Mar 23, 2017 #4

    PeroK

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    Work out ##A## and ##B## and keep going!
     
  6. Mar 23, 2017 #5

    Mark44

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    Surely you don't mean ##\frac A i + 1 + \frac B i - 1##, which is what you wrote.

    If you meant ##\frac A {i + 1} + \frac B {i - 1}##, you need to at least write this as A/(i + 1) + B/(i - 1); i.e., using paretheses.
     
  7. Mar 23, 2017 #6

    vela

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    You shouldn't expect it to because you're comparing apples to oranges. The decomposition is of just 1/(i2-1), but the expression you're trying to derive is for sn. Note that one expression involves the variable ##i## while the other involves ##n##. You have to do a bit more work to connect the two.
     
  8. Mar 23, 2017 #7
    Yes, the stated problem says to find the limit for sn. Can you give me a clue what to do to get on track here ?
     
  9. Mar 23, 2017 #8

    vela

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    Follow PeroK's advice and keep going.
     
  10. Mar 23, 2017 #9
    I did and got A = - 1/2 and B = 1/2 ==> [ - (1/2) / ( i + 1 ) ] + [ (1/2) / ( i - 1 ) ] which does not seem to help. Where is that 3/4 coming from ?
    I do not understand what vela is trying to tell me.
    I know I am way off base here.
     
  11. Mar 24, 2017 #10

    ehild

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    Write out the sum Sn for n=1, n=2, n=3, n=4.... What do you observe?
     
  12. Mar 24, 2017 #11
    OK how about an easier question.
    Why is limit n=> inf ∑ 1 / n2 = π2 / 6 ??

    When I find the Antiderivative I get - 1/x
    Where does the π come from ??
     
  13. Mar 24, 2017 #12

    Mark44

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    This isn't actually an easier problem. Using the advice given in this thread are you able to finish the original problem in this thread?
     
  14. Mar 24, 2017 #13
    No I can not.
    But a p-series is rather basic so if I knew how to do this one
    I will likely induce how to do others. First problem is the hardest one.
    I know I need to work an improper integral, but can not get to π2 / 6
     
  15. Mar 25, 2017 #14

    Mark44

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    I don't think an improper integral is any help in this problem.

    The p-series test tells you whether a series of the form ##\sum_{n = 1}^\infty \frac 1 {n^p}## converges. It doesn't tell you what such a series converges to.

    Have you made any progress on the series you first asked about? That one is an example where you can actually find the value the series converges to.

    The advice already given in this thread is enough to solve the problem you started with. By decomposing ##\frac 1 {i^2 - 1}## I was able to get exactly what it asks you to show.
     
    Last edited: Mar 25, 2017
  16. Mar 25, 2017 #15

    SammyS

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    Similar to @ehild 's advice

    Consider n=5:
    Write out the four terms obtained from ##\displaystyle \ \frac12\frac1{(i-1)}\ ## for ##\ i =\,## 2, 3, 4, 5 . Don't bother to sum these.​

    Similarly,
    Write out the four terms obtained from ##\displaystyle \ \frac{-1}2\frac1{(i+1)}\ ## for ##\ i =\,## 2, 3, 4, 5 . Don't bother to sum these either.​


    Now consider that you need to sum these 8 terms. What pops out at you?.
     
  17. Mar 25, 2017 #16
    Sammy S
    Thank you.
    I can see the 3/4 when I find the 8 terms for n= 2,3,4,5.
    It telescopes after 2 iterations of n.
    And if the second term of the solution in the OP were not - 1/2n but was
    + 1/ 2(n+1) I would have it understood completely. [Maybe the teacher made an error ?]

    I did the Integral Test and got 1/3 not 3/4.
    Can not explain that but now I see that the convergence, if there is one, will
    be the finite value of the integral.
    Still very shaky understanding this but that's what keeps me coming back for more.
    Smile
    Thanks again.
     
  18. Mar 25, 2017 #17

    Mark44

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    No. As written, it is correct. When you expand the summation, write terms for, say, n = 2, 3, 4, 5, as well as terms for n-3, n-2, n-1, and n. You should see that you will have the 3/4 term, but will also have a couple more.
    The integral test just tells you whether the series converges or diverges -- it doesn't give you the sum of the series.
    No. The integral is only loosely related to the summation.
     
  19. Mar 25, 2017 #18
    Thanks Mark44
    Yes if the integral is finite, the series is convergent.
    I did the integral again and got 0.549 as a lower bound
    Adding in f(N) = f(2) = 1/3, the upper bound is 0.882 and 3/4 fall between these bounds.
    This lower/upper bound information is from Wiki ; Integral Series Test.
    Thanks again.
     
  20. Mar 25, 2017 #19

    SammyS

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    Clearly, you are missing the point.

    There is no integral involved.

    Had you done what I suggested you would have gotten the following.

    Consider n=5:
    Write out the four terms obtained from ##\displaystyle \ \frac12\frac1{(i-1)}\,,\ ## for ##i=## 2, 3, 4, 5 .
    ##\displaystyle \frac12, \frac14, \frac16, \frac18 ##​
    Write out the four terms obtained from ##\displaystyle \ -\frac12\frac1{(i+1)}\,,\ ## for ##i=## 2, 3, 4, 5 .
    ##\displaystyle -\frac16, -\frac18, -\frac1{10}, -\frac1{12} ##​

    Some terms cancel when you add these. Which ones don't cancel?
    .
    What happens if n = 12 instead of 5 ?
     
    Last edited: Mar 25, 2017
  21. Mar 26, 2017 #20

    ehild

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    Almost the same as @sammy-s hint...
    The sum Sn can be written as ##S_n=\sum_2^n={\frac{1}{2}\left(\frac{1}{i-1}-\frac{1}{i+1}\right)}=\sum_2^n{a_i}##
    Write up the terms ##a_i=\frac{1}{2}\left(\frac{1}{i-1}-\frac{1}{i+1}\right)## for i=2, 3, 4, 5....
    ##a_2=\frac{1}{2}\left(\frac{1}{2-1}-\frac{1}{2+1}\right)=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)##
    ##a_3=\frac{1}{2}\left(\frac{1}{3-1}-\frac{1}{3+1}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)##
    ##a_4=\frac{1}{2}\left(\frac{1}{4-1}-\frac{1}{4+1}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)##
    ##a_5=\frac{1}{2}\left(\frac{1}{4-1}-\frac{1}{4+1}\right)=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)##
    What is S5, for example? You have to sum the ai-s.
    ##S_5=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3} + \frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}\right)##
    Some terms cancel... Which are the remaining terms? Which are the remaining terms in Sn?
     
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