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Math Help - Infinite Series and Limits

  1. #1
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    Infinite Series and Limits

    This was a problem in my test this morning. I was not able to answer it. But I thought it was pretty interesting, and would like to know the answer. Could you guys help me out?

    \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}
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  2. #2
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    Quote Originally Posted by metallicsatan View Post
    This was a problem in my test this morning. I was not able to answer it. But I thought it was pretty interesting, and would like to know the answer. Could you guys help me out?

    \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}


    \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}=\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2} .

    To what familiar and very easy to integrate function's Riemann sum does the last sum above ressemble?

    And in what interval, of course?

    Tonio
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Last edited by Also sprach Zarathustra; November 10th 2010 at 02:42 AM.
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  4. #4
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    Quote Originally Posted by tonio View Post
    \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}=\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2} .

    To what familiar and very easy to integrate function's Riemann sum does the last sum above ressemble?

    And in what interval, of course?

    Tonio
    That is just the given in the problem. I don't know if it's supposed to be a Riemman sum of some function over an integral.

    Also, is the following statement correct?
    \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2} = <br />
\displaystyle \sum_{n=1}^\infty 1 = \infty
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  5. #5
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    Quote Originally Posted by metallicsatan View Post
    That is just the given in the problem. I don't know if it's supposed to be a Riemman sum of some function over an integral.

    Also, is the following statement correct?
    \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2} = <br />
\displaystyle \sum_{n=1}^\infty 1 = \infty

    No, it isn't correct. And you have to figure out the function, the interval and its partition. Think it hard.

    Tonio
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    I can only think of \frac{1}{1+x^2} but the problem is, the summation does not have a partition, if that is the case. Any more hints?
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    Quote Originally Posted by metallicsatan View Post
    I can only think of \frac{1}{1+x^2} but the problem is, the summation does not have a partition, if that is the case. Any more hints?

    That's the function...now what's the interval and its partition? Check carefully the definition of Riemann sum...

    Tonio
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    so is \Delta x=1? and I=[0,k]. If that is the case, the limit is equal to \int_{0}^{\infty}\frac{1}{1+x^2}\, dx=\frac{\pi}{2}? Right? I'm not sure.
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  9. #9
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    Quote Originally Posted by tonio View Post
    \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}=\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2} .
    This is incorrect. Dividing both numerator and denominator of \frac{n^2}{n^2+ k^2} by n^2 gives \frac{1}{1+ \left(\frac{k}{n}\right)^2}. And, now, unfortunately, that is NOT the same as [/tex]\frac{1}{1+ x^2}[/tex].

    Perhaps dividing both numerator and denominator by k^2 giving \frac{\left(\frac{n}{k}\right)^2}{1+ \left(\frac{n}{k}\right)^2} so we can think of it as a Riemann sum for \int \frac{x^2}{1+ x^2} dx= \int x-\frac{x}{x^2+ 1} will work.

    To what familiar and very easy to integrate function's Riemann sum does the last sum above ressemble?

    And in what interval, of course?

    Tonio
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  10. #10
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    Now I have totally no idea what to do. Any more hints?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    In my poor opinion this problem can be solved without using Riemann sums... ( Thinking in a new direction)
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  12. #12
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    Quote Originally Posted by Also sprach Zarathustra View Post
    In my poor opinion this problem can be solved without using Riemann sums... ( Thinking in a new direction)
    how?
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    Quote Originally Posted by HallsofIvy View Post
    This is incorrect. Dividing both numerator and denominator of \frac{n^2}{n^2+ k^2} by n^2 gives \frac{1}{1+ \left(\frac{k}{n}\right)^2}. And, now, unfortunately, that is NOT the same as [/tex]\frac{1}{1+ x^2}[/tex].

    Perhaps dividing both numerator and denominator by k^2 giving \frac{\left(\frac{n}{k}\right)^2}{1+ \left(\frac{n}{k}\right)^2} so we can think of it as a Riemann sum for \int \frac{x^2}{1+ x^2} dx= \int x-\frac{x}{x^2+ 1} will work.
    using this Riemann integral, what should be the partition and the boundaries?
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  14. #14
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    Quote Originally Posted by HallsofIvy View Post
    This is incorrect. Dividing both numerator and denominator of \frac{n^2}{n^2+ k^2} by n^2 gives \frac{1}{1+ \left(\frac{k}{n}\right)^2}.


    Oops! I totally missed that typo...

    The problem is that k is the one going to infinity...I did a mess with the indexes.

    Oh, well...

    Tonio



    And, now, unfortunately, that is NOT the same as \frac{1}{1+ x^2}.


    Perhaps dividing both numerator and denominator by k^2 giving \frac{\left(\frac{n}{k}\right)^2}{1+ \left(\frac{n}{k}\right)^2} so we can think of it as a Riemann sum for \int \frac{x^2}{1+ x^2} dx= \int x-\frac{x}{x^2+ 1} will work.

    .
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  15. #15
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    Quote Originally Posted by HallsofIvy View Post
    This is incorrect. Dividing both numerator and denominator of \frac{n^2}{n^2+ k^2} by n^2 gives \frac{1}{1+ \left(\frac{k}{n}\right)^2}. And, now, unfortunately, that is NOT the same as [/tex]\frac{1}{1+ x^2}[/tex].

    Perhaps dividing both numerator and denominator by k^2 giving \frac{\left(\frac{n}{k}\right)^2}{1+ \left(\frac{n}{k}\right)^2} so we can think of it as a Riemann sum for \int \frac{x^2}{1+ x^2} dx= \int x-\frac{x}{x^2+ 1} will work.
    Hi, if we change it to riemann sum, should there be kX1/k introduced as well? so we will end up with k\int \frac{x^2}{1+ x^2}dxwhere do the k go from here?
    Last edited by ineedyourhelp; November 10th 2010 at 11:00 PM. Reason: latex typo sorry!
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