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Math Help - sequences

  1. #1
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    sequences

    Could someone just tell me if the following sequences converge or not:

    Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
    For example, if n =2, x =3, then the sequence is (1/3) + (1/4).

    Let bn = the summation of (1/x) beginning with x = n+1 and ending with pn where p is a positive integer greater than one.
    For example, if n =2, x =3, and p =3, then the sequence is (1/3) + (1/4) + (1/5) + (1/6).

    It seems to me that both sequences are increasing and bounded above, which would make them convergent. Is this true or not?
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    Could someone just tell me if the following sequences converge or not:

    Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
    For example, if n =2, x =3, then the sequence is (1/3) + (1/4).

    Let bn = the summation of (1/x) beginning with x = n+1 and ending with pn where p is a positive integer greater than one.
    For example, if n =2, x =3, and p =3, then the sequence is (1/3) + (1/4) + (1/5) + (1/6).

    It seems to me that both sequences are increasing and bounded above, which would make them convergent. Is this true or not?
    The first is as it is of order:

    \int_{n}^{2n}1/x ~dx = \ln(2),

    I will leave it to you to sort out the details of the limit.

    The second I can't tell what the upper limit of summation is supposed to be.

    RonL
    Last edited by CaptainBlack; October 20th 2007 at 12:06 PM.
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  3. #3
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    I'm not sure if you understand what I'm saying.

    I have two sequences:

    the first one is the summation symbol (sigma) with 2n on top and i = (n+1) on the bottom (I changed x to i to eliminate possible confusion), i am summing (1/i) with those conditions

    the second one is the summation symbol (sigma) with pn on top and i = (n+1) on the bottom where p is a positive integer, i am summing (1/i) with those conditions

    are these convergent?
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    Have you noticed that:
    \begin{array}{l}<br />
 a_5  = \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} \\ <br />
 \frac{1}{2} = \frac{5}{{10}} < \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} < \frac{5}{5} = 1 \\ <br />
 \end{array}?

    So we have:
    \frac{1}{2} < a_n  = \sum\limits_{k = n + 1}^{2n} {\frac{1}{k}}  < 1.
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    Quote Originally Posted by PvtBillPilgrim View Post
    I'm not sure if you understand what I'm saying.

    I have two sequences:

    the first one is the summation symbol (sigma) with 2n on top and i = (n+1) on the bottom (I changed x to i to eliminate possible confusion), i am summing (1/i) with those conditions
    That is what I was addressing.

    RonL
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    Quote Originally Posted by PvtBillPilgrim View Post
    the second one is the summation symbol (sigma) with pn on top and i = (n+1) on the bottom where p is a positive integer, i am summing (1/i) with those conditions

    are these convergent?
    \sum_{r=n+1}^{p\times n} 1/r,\ p \in \mathbb{N},\ p \ge 2.

    Yes is convergent to \ln(p) for the same reason that the first
    one converges to \ln(2)

    RonL
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    I have a question. Am I the only one here that has misgivings about the captainís response?
    I have no doubt that that the value of the limit he gave is correct.
    But it seems to me that the question was about showing that the sequence has a limit.
    If I had set this question for a class, I must confess that I think that I would not accept that as an answer.
    I would truly like some input on this question.
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  8. #8
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    Quote Originally Posted by PvtBillPilgrim View Post
    Could someone just tell me if the following sequences converge or not:

    Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
    For example, if n =2, x =3, then the sequence is (1/3) + (1/4).
    \frac{1}{n+1}+...+\frac{1}{2n} = \frac{1}{n}\left( \frac{1}{1+\frac{1}{n}} + ... + \frac{1}{1+\frac{n}{n}} \right) = \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}
    This the the Riemann sum with equal sub-intervals so:
    \lim \ \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}} = \int_0^1 \frac{dx}{x+1} = \ln 2
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  9. #9
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    Quote Originally Posted by Plato View Post
    I have a question. Am I the only one here that has misgivings about the captain’s response?
    I have no doubt that that the value of the limit he gave is correct.
    But it seems to me that the question was about showing that the sequence has a limit.
    If I had set this question for a class, I must confess that I think that I would not accept that as an answer.
    I would truly like some input on this question.
    If you read what I wrote I was leaving it to the poster to work out the detail.

    The idea is that:

    <br />
\sum\limits_{k = n + 1}^{2n} {\frac{1}{k}} = \int_n^{2n} 1/x ~dx + O(1/n)<br />

    With some effort you can establish explicit bounds on the error, which is
    what I want the poster to do if he cannot see that it is O(1/n) which
    is all we need to establish what the limit actually is.

    Think of this as an illustration of how we really do these things.

    RonL
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