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Find an expression for the nth partial sum using this identity

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data

    The series,

    1 / (1 x 2) + 1 / (2 x 3) + 1 / (3 x 4) + ... + 1 / [n(n + 1)] + ...

    is not a geometric series.

    (A) Use the identity

    1 / [k(k + 1)] = 1 / k - 1 / (k + 1)

    to find an expression for the nth partial sum Sn and (B) use it to find the sum of the series. (C) Also, how many terms of the series would be required so that the partial sum differs from the total sum by less than 0.001?

    2. Relevant equations

    1 / [k(k + 1)] = 1 / k - 1 / (k + 1)

    3. The attempt at a solution

    (A)

    I do not understand how the provided identity could be used to find Sn.
    However that I think that the following is true,

    Sn = 1 / (1 x 2) + 1 / (2 x 3) + ... + 1 / [n(n + 1)]

    I cannot seem to make an expression such as the answer, 1 - 1 / (n +1).

    (B)

    I understand that the limit of Sn as n approaches infinite is equal to the sum of the series, but, alas, I do not have an expression for Sn. (The answer is 1.)

    (C)

    From my incomplete understanding of infinite series, I would think that the partial sum Sn would have a difference of less than 0.001 with the total sum L (the limit of the infinite series), or,

    Sn - L < 0.001
    Sn < 0.001 + L

    Here, I would think there would be manipulation of Sn to isolate n for this inequality. (The answer is 1000.)


    ----

    Much appreciation for any help! This is my first time in maybe 3 years since doing any form of obvious sequences and series, so the smallest of corrections (incorrect terminology) is completely welcomed.
     
  2. jcsd
  3. Jul 14, 2011 #2
    How can you use the given identity to represent 1/(1x2)? 1/(2x3)? And every term in the series?
     
  4. Jul 15, 2011 #3
    Hmmmm...

    S1 = 1 / (1 x 2) = 1 / [1(1 + 1)] = 1 / 1 - 1 / (1 + 1) = 1 - 1 / 2 = 1 / 2
    S2 = 1 / (2 x 3) = 1 / [2(2 + 1)] = 1 / 2 - 1 / (2 + 1) = 1 / 2 - 1 / 3 = 1 / 6
    S3 = 1 / (3 x 4) = 1 / [3(3 + 1)] = 1 / 3 - 1 / (3 + 1) = 1 / 3 - 1 / 4 = 1 / 12

    .
    .
    .

    Sn = S1 + S2 + ... + 1 / n - 1 / (n + 1)

    And, as I see adding S1, S2, and so forth, the limit as n approaches infinite of is equal to 1 (it appears). So...


    Sn = 1 + 1 / n - 1 / (n + 1)


    And, as I understand that the limit of 1 / n as n approaches infinite is equal to 0...

    Sn = 1 + 0 - 1 / (n + 1) = 1 - 1 / (n + 1)


    Is this correct? Mind you, I found the number 1 by manual calculations (I don't know if there is a 'more correct' method).
     
  5. Jul 15, 2011 #4
    Instead of evaluating the fractions, leave them as they are. Then you can write the partial sum as:

    (1-1/2)+(1/2-1/3)+(1/3-1/4)+....

    Can you use this form now?
     
  6. Jul 18, 2011 #5
    I can now see the validity of the answer 1 - 1 / (n + 1) (2)
    as the nth partial sum, in that for the first term it is immediately obvious, and that in using (2) for the nth term its difference can be verified by adding the previous partial sums.

    However, I recognized this by looking at the first term and seeing the solution (2), and by applying it to an nth term and then adding the partial sums before it and seeing that they are equal. Was this what I was supposed to do?
     
  7. Jul 18, 2011 #6
    I'm not entirely sure what you are asking; what do you mean by "previous partial sums"? The nth partial sum is the (n-1)th partial sum plus the nth term by definition. You were supposed to see the validity of the nth partial sum based on the hint I think.
     
  8. Jul 19, 2011 #7
    To be honest, I not entirely sure what I am doing, and am still pretty confused.

    But, what I meant but "previous partial sums" is that in using (2) for n = 1, it results in the first value of 1 / 2; for the second term n = 2, the equation gets the same value as in adding the first and second terms; and so on.
     
  9. Jul 19, 2011 #8

    vela

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    You're confusing the two sequences. You have one sequence[tex]a_k = \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}[/tex]and the sequence of partial sums[tex]S_n = \sum_{k=1}^n a_k=a_1 + a_2 + \cdots + a_n[/tex]Calculate the n-th partial sum Sn by substituting in for a1, a2, …, an and using the hint. (I think this is what you may have done, but it's not clear because you're using Sn to refer to both sequences.)
     
  10. Jul 19, 2011 #9
    Yes, that is what the partial sum represents; the sum of the first n terms of a sequence. Of course, that is what you get when you evaluated-

    (1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-1/(n-1)) = 1- 1/(n-1)
     
  11. Jul 30, 2011 #10
    Well, I appreciate all the responses, but I have become a bit frustrated and am going to give this problem some time and come back to it later, and maybe I will understand these hints and help I have been given a bit better, haha.
     
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