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Thread: 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:

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

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

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

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

    Subtract your 1 and you have 1 as your answer.

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

    This is the same as:

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

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

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


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

    Here's an elementary approach . . .


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

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

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

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


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

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

    Thanks
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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 $\displaystyle \frac{n}{2^{n-1}}$. The trick is to write down a function $\displaystyle f(x)$ for which the sum is a special value. Here we consider $\displaystyle f(x) = \sum_n \frac{n}{x^{n-1}}$, so that you want $\displaystyle f(2)$. We look at $\displaystyle x^{-2} f(x) = \sum_n \frac{n}{x^{n+1}}$ in order to be able to integrate term by term. So $\displaystyle x^{-2}f(x)$ is the derivative of $\displaystyle \sum_n \frac{-1}{x^n}$, which is $\displaystyle \frac{-1}{1-1/x} = \frac{x}{1-x} = 1+\frac{1}{1-x}$. Hence $\displaystyle x^{-2} f(x) = \frac{1}{(1-x)^2}$ and your sum is $\displaystyle f(2) = 4$.
    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
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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:


    $\displaystyle
    S(c,a,b)=\sum_{n=a}^b n.c^{n-1}
    $

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

    $\displaystyle
    S(x,a,b)=\sum_{n=a}^b n.x^{n-1}
    $

    each term of this is obviously the derivative wrt $\displaystyle x$of:

    $\displaystyle
    t(x,n)=x^n
    $.

    So we can integrate term by term to get:

    $\displaystyle
    I(x,a,b)=\sum_{n=a}^b x^{n}
    $

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

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

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

    and:

    $\displaystyle
    S(c,a,b)=\left{} \frac{d}{dx}I(x,a,b) \right|_{x=c}
    $

    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; Jul 9th 2006 at 09:49 PM.
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