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Math Help - Convergent infinite series

  1. #1
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    Convergent infinite series

    Ok
    So I have the problem of:
    A definition of a convergent infinite series with the sum S is:
    Given any small positive number (ε) it possible to find an integeer N so that |S-Sn|<ε for every n>N

    Select some ε's and find the corresponding N's for the following series:


    This is how far I got:



    so

    Let ε=1/8

    Then |S-Sn|<1/8 for some n>N
    so |1-Sn|<1/8
    sn will always be positive and less than S ie less than 1 so
    1-Sn<1/8
    Sn>7/8

    Now I am stuck... am I doing this right? If so how do I find N?

    Any help appreachated.

    colgon
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  2. #2
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    You want the difference between your sum (through some N) and the number the series converges to to be less than epsilon.

    Let N=3, then your sum is 1 + (1/2) + (1/4) + (1/8) = 1.875

    |1.875-2| = .125 = (1/8) (Noting that the series converges to 2)

    So at N=3, the difference = epsilon. Therefore, for n > N, the difference will be less than epsilon=(1/8).

    Now just pick some more epsilons and see how many terms it takes to get a difference less than it (your chosen epsilon that is!)


    Hope that helps.
    Last edited by boomlegboom; April 16th 2009 at 05:58 PM.
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  3. #3
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    thanks

    Thank you,
    I'll give it a go.

    EDIT
    ok looking at this I don't understand why the sum of the series converges to 2, I thought it was 1?

    Also have you changed mod [sum of series - partial sum] with mod of [partial sum - mod of series]? EDIT ok its a mod it don't matter

    Wait I see, you have 1 + (1/2) + (1/4) + (1/8) = 1.875
    The sum is from n=1 so I think the first term in the series is 1/2 not 1?
    Last edited by colgon; April 16th 2009 at 07:27 PM.
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  4. #4
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    Question is this it?

    ok I think i got it.

    Assuming the sum of the series is 1 (as shown in original post, please explain to me if I am mistaken.

    Then leading on from where the original post left off at Sn>7/8

    We are looking for N to satify Sn>7/8 for some N where n>N

    Using Sn=a(1-r^n)/(1-r)

    0.875<0.5(1-0.5^n)/(1-0.5)
    and
    0.875<0.5(1-0.5^n)/0.5
    and
    0.875<1-0.5^n
    and
    -0.125<-0.5^n
    and
    0.125<0.5^n
    and
    ln0.125<nln0.5
    and
    (ln0.125)/ln0.5<n
    and
    3<n

    So N=3

    EDIT EDIT my inequality manipulation is wrong on this post as I have divided though by -1?

    Also can anyone tell me if my mod and inequalitiy manipulation in the original post is correct????

    Soz this is a mess... it must be late and time for bed, if you have managed to read though all this an you have any answers please let me know.
    Last edited by colgon; April 16th 2009 at 07:37 PM.
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  5. #5
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    My mistake, didn't notice that the sum started at n=1..

    Sn is the sum of the series up to the nth term, so

    S1 = 1/2
    S2 = 1/2 + 1/4 = .75
    S3 = 1/2 + 1/4 + 1/8 = .875
    ...and so on.

    The definition of convergence says that for any ε (so someone could pick it as small as they want to) there exists an N (integer: 1,2,3,..) such that for n>N |S-Sn|<ε. So if they said ε = 1, |S-S1|=|1 -1/2|=1/2<1=ε. So N=1, and for any n>1, |S-Sn|<ε is still true.

    Now, say ε = 1/100. Your job (to prove convergence) is to find the n such that |1-Sn|<1/100.

    Sn > 1-1/100 = 0.99

    So, S7 = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 = 0.9921875 > 0.99

    Therefore, N>=7 satisfies |1-Sn|<1/100.

    The question really just boils down to how many (N=?) terms of the series are necessary before the sum is within ε of S=1.
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  6. #6
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    This is a geometric sum, so you can calculate the n^{th} term, EVEN if the infinite series doesn't converge.

    Let S_n=\sum_{k=0}^n a^k.

    Then S_n=1+\cdots +a^n and {S_n-1\over a}+ a^n=S_n when a is not zero.

    Thus S_n(1-a)=1-a^{n+1}, giving us S_n={1-a^{n+1}\over 1-a}.

    NOW, if -1<a<1 the infinite series, which is the limit of the above series, converges to {1\over 1-a}.
    Last edited by matheagle; April 17th 2009 at 07:34 AM.
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  7. #7
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    Question So i think i have it

    LEt ε=1/100
    So
    |S-Sn|<1/100
    ⇒|1-Sn|<1/100
    As 0<Sn<1 execpt at limit we can say
    1-Sn<1/100
    ⇒-Sn<-99/100
    ⇒Sn>99/100

    Now given Sn=a(1-r^n)/(1-r)

    ⇒0.5(1-0.5^n)/0.5>99/100
    ⇒1-0.5^n>99/100
    ⇒-0.5^n>-1/100
    ⇒0.5^n<1/100
    ⇒nln0.5<ln1/100
    ⇒n>(ln1/100)/ln0.5
    ⇒n>6.64...

    If N ∈ ℤ
    then to satisfy n>N

    N=6

    Correct??
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  8. #8
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    Since n>6.64, the next greatest integer is 7, so that would be your N.
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  9. #9
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    n>=7
    or
    n>6 ?
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  10. #10
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    the definition says for n>=N, so n>=7.
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