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Math Help - compute a sum

  1. #1
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    compute a sum

    hello

    i need to compute this sum:

    sigma(n=1 to infinity) n^2 / 2^n

    and i'm having trouble doing that.
    i tried looking at a power series,but couldn't develop it.

    help would be appreciated.
    thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by parallel
    hello

    i need to compute this sum:

    sigma(n=1 to infinity) n^2 / 2^n

    and i'm having trouble doing that.
    i tried looking at a power series,but couldn't develop it.

    help would be appreciated.
    thanks.
    This is a bit cryptic, but if you can do it from this you will have done
    a lot of the work yourself. If you have trouble we will fill in the detail.

    Consider the geometric series:

    <br />
S(x)=\sum_1^{\infty} (x/2)^n<br />

    which you should be able to sum for |x|<2. Now differentiate
    once and twice and see what you find after a bit of rearrangement of the
    corresponding sums of the series, and letting x=1.

    RonL
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  3. #3
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    Hello, parallel!

    I just "invented" an approach to this problem . . .


    \sum^{\infty}_{n=1}\frac{n^2}{2^n}


    We have: . S\;=\;\frac{1}{2} + \frac{4}{2^2} + \frac{9}{2^3} + \frac{16}{2^4} + \frac{25}{32} + \hdots [1]

    Multiply by \frac{1}{2}:

    . . . . . . . \frac{1}{2}S \;= \qquad\frac{1}{2^2} + \frac{4}{2^3} + \frac{9}{2^4} + \frac{16}{2^5} + \hdots [2]

    Subtract [2] from [1]:

    . . . . . . . \frac{1}{2}S\;=\;\frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \frac{7}{2^4} + \frac{9}{2^5} + \hdots [3]

    Multiply by \frac{1}{2}:

    . . . . . . . \frac{1}{4}S\;= \qquad\frac{1}{2^2} + \frac{3}{2^3} + \frac{5}{2^4} + \frac{7}{2^5} + \hdots [4]

    Subtract [4] from [3]:

    . . . . . . . \frac{1}{4}S\;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{2}{2^3} + \frac{2}{2^4} + \frac{2}{2^5} + \hdots

    . . . . . . . \frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}
    . . . . . . . . . . . . . . an infinite geometric series with a = \frac{1}{2},\;r = \frac{1}{2}
    . . . . . . . . . . . . . . . . . . Its sum is: . \frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1

    Hence, we have: . \frac{1}{4}S \;= \;1 + \frac{1}{2} \;= \;\frac{3}{2} . . . . (drumroll, please)

    Therefore: . \boxed{S\;=\;6}
    . . . . ta-DAA!

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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Soroban
    Hello, parallel!

    I just "invented" an approach to this problem . . .




    We have: . S\;=\;\frac{1}{2} + \frac{4}{2^2} + \frac{9}{2^3} + \frac{16}{2^4} + \frac{25}{32} + \hdots [1]

    Multiply by \frac{1}{2}:

    . . . . . . . \frac{1}{2}S \;= \qquad\frac{1}{2^2} + \frac{4}{2^3} + \frac{9}{2^4} + \frac{16}{2^5} + \hdots [2]

    Subtract [2] from [1]:

    . . . . . . . \frac{1}{2}S\;=\;\frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \frac{7}{2^4} + \frac{9}{2^5} + \hdots [3]

    Multiply by \frac{1}{2}:

    . . . . . . . \frac{1}{4}S\;= \qquad\frac{1}{2^2} + \frac{3}{2^3} + \frac{5}{2^4} + \frac{7}{2^5} + \hdots [4]

    Subtract [4] from [3]:

    . . . . . . . \frac{1}{4}S\;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{2}{2^3} + \frac{2}{2^4} + \frac{2}{2^5} + \hdots

    . . . . . . . \frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}
    . . . . . . . . . . . . . . an infinite geometric series with a = \frac{1}{2},\;r = \frac{1}{2}
    . . . . . . . . . . . . . . . . . . Its sum is: . \frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1

    Hence, we have: . \frac{1}{4}S \;= \;1 + \frac{1}{2} \;= \;\frac{3}{2} . . . . (drumroll, please)

    Therefore: . \boxed{S\;=\;6}
    . . . . ta-DAA!


    This is implicit in the method I outlined

    CB
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  5. #5
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    thanks for the reply CaptainBlack,this is what I came up with,hope it's correct

    the sum of (x/2)^n is : x/(2-x)

    taking deriv:
    (1) sigma n/(2^n)*x^(n-1) = 2/(2-x)^2

    taking 2nd deriv:

    sigma n(n-1)/2^n * x^(n-2)
    this sum is:

    (2) n^2/2^n*x^(n-2) - n/2^n*x^(n-2) = deriv of (1) = (8-4x)/(2-x)^4

    now I noticed that the second term is (1) * (1/x),
    so (2) + (1)*(1/x) yields the result,which is (4x+4x-2x^2)/(2-x)^3 = 6 (if x=1)

    is this correct?
    thanks for the help

    Soroban,
    hey, I really liked your solution,I'll sure keep this approach in mind
    thanks for your help too.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by parallel
    thanks for the reply CaptainBlack,this is what I came up with,hope it's correct

    the sum of (x/2)^n is : x/(2-x)

    taking deriv:
    (1) sigma n/(2^n)*x^(n-1) = 2/(2-x)^2

    taking 2nd deriv:

    sigma n(n-1)/2^n * x^(n-2)
    this sum is:

    (2) n^2/2^n*x^(n-2) - n/2^n*x^(n-2) = deriv of (1) = (8-4x)/(2-x)^4

    now I noticed that the second term is (1) * (1/x),
    so (2) + (1)*(1/x) yields the result,which is (4x+4x-2x^2)/(2-x)^3 = 6 (if x=1)

    is this correct?
    thanks for the help

    Soroban,
    hey, I really liked your solution,I'll sure keep this approach in mind
    thanks for your help too.
    That's what I had in mind. I will not check it in detail as the answer agrees
    with Soroban's so we may presume the algebra is correct.

    This method is powerful as it provides a method of finding:

    <br />
S=\sum_1^{\infty}P(n)/k^n<br />

    for any polynomial P(n) in n (at least for
    smallish polynomials) and k>1.

    RonL
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  7. #7
    Junior Member
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    thanks again
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