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Difficult Series Question

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  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem says approximate the sum to within 1/100 of the value of the infinite sum, and the sum is
    1/2-(1x3)/(2x4)+(1x3x5)/(2x4x6) and so on... (Teacher said I can leave answer in summation notation so I just need to find how many terms I need to add together.)

    2. Relevant equations
    n/a

    3. The attempt at a solution

    Since it is an alternating series, I would just need to find the first term that is equal to or less than 1/100 to know how many of the terms I need to add together and the terms are (1x3x...x(2n+1))/(2x4x...(2n+2)). However, I cannot think of any way at all to find when one of these individual terms is equal to or less than 1/100.

    My goal right now is finding the first term that will equal 1/100 basically... but I do not see any way this is possible. If there is a way I can find it please help me; if I should be going a completely different way to solve this problem please let me know. I've been thinking about this problem for a few hours and just cannot come up with anyway of finding the answer. Thanks for any help!
     
    Last edited by a moderator: Oct 13, 2015
  2. jcsd
  3. Oct 13, 2015 #2

    Mark44

    Staff: Mentor

    Can you write the general term of the series? If so, use it to figure out what n needs to be so that the general term is < .01.
     
  4. Oct 13, 2015 #3
    If you write ##(2*4*6*\dots) = 2(1*2*3*\dots)## Is there any terms that cancel each other?
     
  5. Oct 13, 2015 #4
    @Mark44 The General Term of the series would just be (-1)^n(1x3x...x(2n+1))/(2x4x...x(2n+2)!) but I can't really use this algebraically to find the value for n
     
  6. Oct 13, 2015 #5

    Mark44

    Staff: Mentor

    No, that's not what it is. You shouldn't have that factorial in there. ##2 \cdot 4 \cdot 6 \dots \cdot (2n + 2) = 2(n + 1)!##. Maybe that's what you were shooting for.

    Anyway, write your inequality (can omit the ##(-1)^n##) and solve. Typically the way to solve inequalities like this one is to try a few values of n to see when you get a true statement.
     
  7. Oct 13, 2015 #6
    @Mark44 I just accidently put in the ! I didn't mean to
     
  8. Oct 13, 2015 #7
    Also... 2*4*6*8 doesnt equal 2(n+1)! it would equal 2^n(n+1)!
     
  9. Oct 13, 2015 #8

    Mark44

    Staff: Mentor

    Yes, my mistake -- you are right, almost. 2⋅4⋅6⋯⋅(2n+2)=2n + 1(n+1)!
     
  10. Oct 13, 2015 #9
    but I can rewrite 2x4x6x...(2n+1) as 2^n(n!) but rewriting 1x3x5... seems a little more difficult
     
  11. Oct 13, 2015 #10
    should probably be 2^(n+1)(n+1)! for 2x4x6... my mistake
     
  12. Oct 14, 2015 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Computationally, it might be easiest to not try to get a general formula and then figure out the value of ##n## from that. It might be more straightforward to write the series as
    [tex] S = t_1 - t_2 + t_3 - \cdots [/tex]
    where
    [tex] t_1 = \frac{1}{2}\; \; \text{and}\; \; t_n = \frac{2n-1}{2n} t_{n-1}, \; n \geq 2 [/tex]
    You can just keep calculating the ##t_n## until ##t_n < 0.01## (for example, in a spreadsheet). Doing that is also useful in revealing just how slowly the series converges, and how many terms you need to keep to have any kind of decent accuracy. For example, 1000 terms are not nearly enough to give you ##t_n < 0.01##.

    However, if you really want a formula for ##t_n## you can apply the result
    [tex] 2 \times 4 \times 6 \times \cdots \times 2k = 2^k k! [/tex]
    to the formula
    [tex] t_n = \frac{1 \times 3 \times \cdots \times 2n-1}{2 \times 4 \times \cdots \times 2n}
    = \frac{(2n-1)!}{(2 \times 4 \times \cdots \times (2n-2) )(2 \times 4 \times \cdots \times 2n)} [/tex]
    This will give you a formula involving factorials and powers of 2 only.
     
    Last edited: Oct 14, 2015
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