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Math Help - Evaluate the summation

  1. #1
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    Evaluate the summation

    Evaluate:

    from 1 to infinity the summation of [n/2^n-1]

    I tried to work it out for a small amounts of n to see if I can see anything happening. What should I do next?
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  2. #2
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    Quote Originally Posted by Nichelle14
    Evaluate:

    from 1 to infinity the summation of [n/2^n-1]

    I tried to work it out for a small amounts of n to see if I can see anything happening. What should I do next?
    Hello, Nichelle,

    I'm puzzled, because it is not clear for me, what you mean. Do you mean:

    \frac{n}{2^n}-1\quad \mbox{or}\quad \frac{n}{2^n-1}\quad \mbox{or}\quad \frac{n}{2^{n-1}}

    So sorry I can't help you.

    Greetings

    EB
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  3. #3
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    Here's a general technique that may help. I'll tackle the example \frac{n}{2^{n-1}}. The trick is to write down a function f(x) for which the sum is a special value. Here we consider f(x) = \sum_n \frac{n}{x^{n-1}}, so that you want f(2). We look at x^{-2} f(x) = \sum_n \frac{n}{x^{n+1}} in order to be able to integrate term by term. So x^{-2}f(x) is the derivative of \sum_n \frac{-1}{x^n}, which is \frac{-1}{1-1/x} = \frac{x}{1-x} = 1+\frac{1}{1-x}. Hence x^{-2} f(x) = \frac{1}{(1-x)^2} and your sum is f(2) = 4.
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  4. #4
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    I'll try the \sum_{n=1}^{\infty}\frac{n}{2^{n}}-1

    You can just use the closed form for a geometric series, namely, \frac{1}{1-x}

    \frac{1}{1-\frac{1}{2}}=2

    Subtract your 1 and you have 1 as your answer.

    \sum_{n=1}^{\infty}\frac{n}{2^{n-1}} can be treated similarly.

    This is the same as:

    \frac{n}{2^{n}2^{-1}}

    2\sum_{n=1}^{\infty}\frac{n}{2^{n}}=\frac{2}{1-\frac{1}{2}}=4
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  5. #5
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    Hello, Nichelle14!

    I'll assume that the exponent is n-1.


    Evaluate: . \sum^{\infty}_{n=1}\frac{n}{2^{n-1}}

    Here's an elementary approach . . .


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

    Divide by 2: . \frac{1}{2}S \;= \;\;\;\,\quad\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \hdots

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

    The right side is a geometric series with a = 1,\;r = \frac{1}{2}
    . . Hence, its sum is: . \frac{1}{1 - \frac{1}{2}} \,= \,2


    Therefore, we have: . \frac{1}{2}S\:=\:2\quad\Rightarrow\quad \boxed{S\,=\,4}

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  6. #6
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    It is the third choice. n-1 is an exponent of 2.

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  7. #7
    Super Member malaygoel's Avatar
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    Quote Originally Posted by rgep
    Here's a general technique that may help. I'll tackle the example \frac{n}{2^{n-1}}. The trick is to write down a function f(x) for which the sum is a special value. Here we consider f(x) = \sum_n \frac{n}{x^{n-1}}, so that you want f(2). We look at x^{-2} f(x) = \sum_n \frac{n}{x^{n+1}} in order to be able to integrate term by term. So x^{-2}f(x) is the derivative of \sum_n \frac{-1}{x^n}, which is \frac{-1}{1-1/x} = \frac{x}{1-x} = 1+\frac{1}{1-x}. Hence x^{-2} f(x) = \frac{1}{(1-x)^2} and your sum is f(2) = 4.
    Rgep, Will you please explain the use of integration and derivative here. I am not able to get it.

    Keep Smiling
    Malay
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by malaygoel
    Rgep, Will you please explain the use of integration and derivative here. I am not able to get it.

    Keep Smiling
    Malay
    When you have a series of the form:


    <br />
S(c,a,b)=\sum_{n=a}^b n.c^{n-1}<br />

    with a,\ b,\ and c numerical constants,
    we consider the function:

    <br />
S(x,a,b)=\sum_{n=a}^b n.x^{n-1}<br />

    each term of this is obviously the derivative wrt xof:

    <br />
t(x,n)=x^n<br />
.

    So we can integrate term by term to get:

    <br />
I(x,a,b)=\sum_{n=a}^b x^{n}<br />

    Now if RHS is a geometric series and may be summed:

    I(x,a,b)=x^a \frac{1-x^{b-a+1}}{1-x},

    (you will need to check this last sum, I did it in a bit of a rush )

    and:

    <br />
S(c,a,b)=\left{} \frac{d}{dx}I(x,a,b) \right|_{x=c}<br />

    Now you need only play around with the values of a, b and c (and the odd
    limiting process to get the required result).

    This is quite a common trick employed to find the sum of series.

    RonL
    Last edited by CaptainBlack; July 9th 2006 at 09:49 PM.
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