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Math Help - Evaluate a Riemann sum

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    Evaluate a Riemann sum

    Evaluate lim(n->inf) \sum^n_{k=1}\frac{(n+2k)^2}{n^3} and prove your result.
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    Quote Originally Posted by tttcomrader View Post
    Evaluate lim(n->inf) \sum^n_{k=1}\frac{(n+2k)^2}{n^3} and prove your result.
    Your summand is:
    \frac{n^2 + 4kn + 4k^2}{n^3} = \frac{1}{n} + \frac{4k}{n^2} + \frac{4k^2}{n^3}

    Now, of all of these the 1/n sum is going to be the most important because the sum of the other two as n approaches infinity converges, whereas the 1/n series does not. So your series does not converge.

    There are innumerable ways to prove that the 1/n series does not converge and that the other two do so I leave this part of the proof to you.

    -Dan
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    Quote Originally Posted by tttcomrader View Post
    Evaluate lim(n->inf) \sum^n_{k=1}\frac{(n+2k)^2}{n^3} and prove your result.
    Evaluate:

    Lim(n -> infty) [Sum(k=1..n) (n+2k)^2 / n^3]

    First look at:

    S(n)= Sum(k=1..n) (n+2k)^2 / n^3 = Sum(k=1..n) (n^2+4kn+4k^2) / n^3

    .....=[Sum(k=1..n)1/n] + (4/n^2)[Sum(k=1..n)k] + (4/n^3)[Sum(k=1..n)(k^2)]

    .....=1 + (4/n^2) n(n+1)/2 + (4/n^3) n(n+1)(2n+1)/6

    .....= 1 + (2+2/n) + (4/3 +2/n +(2/3)/n^2)

    Hence:

    Lim(n -> infty) [Sum(k=1..n) (n+2k)^2 / n^3]=Lim(n -> infty) S(n) = 13/3


    (But I would recomend that you check my working!)

    RonL
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    Quote Originally Posted by topsquark View Post
    Your summand is:
    \frac{n^2 + 4kn + 4k^2}{n^3} = \frac{1}{n} + \frac{4k}{n^2} + \frac{4k^2}{n^3}

    Now, of all of these the 1/n sum is going to be the most important because the sum of the other two as n approaches infinity converges, whereas the 1/n series does not. So your series does not converge.

    There are innumerable ways to prove that the 1/n series does not converge and that the other two do so I leave this part of the proof to you.

    -Dan
    But still it does

    (did it really take me most of an hour to type this? Judging by the time stamps on my post and
    topsquarks it must have taken one of us fifty odd mins.)

    RonL
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    But still it does

    (did it really take me most of an hour to type this? Judging by the time stamps on my post and
    topsquarks it must have taken one of us fifty odd mins.)

    RonL
    Ah, I've run into this SNAFU before. I forgot we are considering the sum, not just the individual terms.

    -Dan
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