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Math Help - Limit of a series.

  1. #1
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    Exclamation Limit of a series.

    What is the limit (if there is one) of the series of numbers? Start by making a table for each series and calculate the value of the series, as more terms are included in the series
    1. 1-1/3+1/5-1/7+1/9
    Number of terms added Sum of these terms
    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    2. 1+ 1/1+1/(1*2)+1/(1*2*3) + 1/(4!)
    3. 1+++1/8
    4. 1+ 1/(2^2) + 1/(3^2) + 1/(4^2)
    5. 2/1*2/3*4/3*4/5*6/5*6/7*8/7*8/9
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  2. #2
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    First, find the general form of your series.

    For 1: \sum_{k=0}^n \frac{(-1)^n}{1+2n}

    So we know that \sum_{k=0}^n \frac{(-1)^n}{1+2n} < \sum_{k=0}^n \frac{(-1)^n}{2n} = \frac{1}{2}\sum_{k=0}^n\frac{(-1)^n}{n}

    By the comparison theorem, we know that the first series converges if the second series converges. By the theorems about summation by parts, we know that the second series converges (that is quite a long proof which is why i do not write it down, should be in every introductury analysis book, however). You can derive your values by setting n = k + 5 where 5 are the values you already have given and k is your row in the table.

    If I remember it correctly the result of this series is \frac{\pi^2}{6} but it is not that easy to show it.
    Last edited by raphw; October 13th 2010 at 02:34 PM.
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  3. #3
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    Quote Originally Posted by raphw View Post
    First, find the general form of your series.
    For 1: \sum_{k=0}^n \frac{(-1)^n}{1+2n}
    So we know that \sum_{k=0}^n \frac{(-1)^n}{1+2n} < \sum_{k=0}^n \frac{(-1)^n}{2n} = \frac{1}{2}\sum_{k=0}^n\frac{(-1)^n}{n}
    By the comparison theorem,
    @raphw;570359, The comparison cannot be used here because this is an alternating series.
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  4. #4
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    for number two you have:

    s_n = \sum_{k=0}^n \frac{1}{n!}

    You can show that \frac{1}{n!} < \frac{1}{2^{n-1}} for n \ge 2.

    Thus: s_n = \sum_{k=0}^n \frac{1}{n!} < 1+ \sum_{k=0}^n \frac{1}{2^n} = 1+ \sum_{k=0}^n {(\frac{1}{2})}^n

    The later converges by the principles of gemetric series and will be for n \to \infty: \frac{1}{1-\frac{1}{2}} = 2.

    Thus \forall n > 2: s_n = \sum_{k=0}^n \frac{1}{n!} < 1 + 2 = 3 and thus your series is bounded above. Since s_n < s_{n+1} \forall n your series is also monotonocally increasing so your series must converge. You might also know that it is this series how the Eulerian number (English?) is most commonly defined.
    Last edited by raphw; October 13th 2010 at 02:36 PM.
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  5. #5
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    Quote Originally Posted by Plato View Post
    @raphw;570359, The comparison cannot be used here because this is an alternating series.
    true, my mistake. However, it is even easier. Since the absolute values of the sequence the series is based on decrease, and the elements are alternating between negative and positive values and the sequence value converges to zero, the summation by parts theorems are already applyable. (By that theorem I mean Leibnitz)
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  6. #6
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    Quote Originally Posted by raphw View Post
    true, my mistake. However, it is even easier. Since the absolute values of the sequence the series is based on decrease, the summation by parts theorems are already applyable. (By that I mean Leibnitz)
    OK
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  7. #7
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    For 3 you have: s_n = \sum_{k=0}^n {(\frac{1}{2})}^n. All the information needed to solve this one is given above, so you might wanna try to solve this by yourself.

    Number 4: There is a theorem that say that if a sequence is stricly positive and the elements of the sequence that your series is based on strictly decrease, then

    \sum_{n=0}^{\inf} a_n converges if and only if \sum_{k=0}^{\inf} 2^k a_{2^k} converges. Prove the later and you are done.
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  8. #8
    Senior Member yeKciM's Avatar
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    where do you have problem with this ? what do you don't understand ?
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