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Error estimate for alterning series

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the smallest integer n that ensures that the partial sum [tex] s_n [/tex] approximates the sum s of the series with error less than 0.001 in absoulute value.

    [tex] \sum_{n=1}^\infty (-1)^{(n+1)}\frac{n}{2^n} [/tex]


    2. Relevant equations

    [tex] {\abs{s-s_n}} <= s_{(n+1)} = a_{(n+1)}[/tex]

    (How can I do absolute value in LaTex?)



    3. The attempt at a solution

    I've got this far.

    error <= first term = [tex] \frac{n+1}{2^{n+1}} [/tex]

    (because (-1)^n+1 doesn't matter in this example)

    [tex] s - s_n < \frac{n+1}{2^n+1} < 0.001 [/tex]

    Here I get lost.
    From example in my textbook which says

    [tex] a_n = \frac{1}{1+2^{n+1}} [/tex]

    This error is less than 0.001 if [tex] 1+2^{(n+1)} > 0.001 [/tex] Since [tex] 2^{10} = 1024, n+1 = 10 [/tex] will do. Hence n = 9.

    In my example I have n on both sides of the fraction. So what do I do?
     
  2. jcsd
  3. Nov 22, 2007 #2
    Since n has to be an integer, I would say just brute force it.
     
  4. Nov 22, 2007 #3
    Brute force it? Do you mean just check in my calculator ? If you are referring to that, it doesn't help me since I'm not allowed to have anything in my test but a pencil and a paper.
     
  5. Nov 22, 2007 #4
    I wouldn't expect to be able to solve this analytically. All I can say is that I know it exists: (n+1)/2^{n+1} tends to zero monotonically.
     
  6. Nov 23, 2007 #5
    Well, I got the answer n = 9 by doing exactly as the example from the textbook but the answer is n = 13.

    What does analytical and monotonical mean ?
     
  7. Nov 23, 2007 #6
    I got n = 13 via brute force.

    Edit: In all fairness, it's not that hard to calculate the term when n = 13; the powers of 2 are pretty easy. I reckon any test question similar to this would have a more simple question in store.
     
  8. Nov 23, 2007 #7
    Yeah but you see Kreizhn I'm not allowed to use calculator so I have to use some method and solve the problem by hand. Sure, I can check if the answer is correct with my calculator but that doesn't help me when I get to the test where i'm not allowed to have anything but a pencil.

    The problem for me is Algebra. I must have slept over or skipped some lectures few years ago when some Algebra rules were being tought.

    I don't have any problem understanding the example from the book but I don't know how to find n in my problem. That's my question.

    EDIT: It's true, the powers of 2 are really easy but when it comes to 6 or 8 or 112 what then? That is the reason I have to find some method finding n
     
  9. Nov 23, 2007 #8
    That is my point though; as ZioX had said, finding the solution to the previous problem is very difficult to do analytically even with a CAS. If you get a similar question on a test where you're not allowed to use a calculater, it's likely that the problem will be simplified such that you won't need one.
     
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