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Math Help - Series

  1. #1
    Member srirahulan's Avatar
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    Lightbulb Series

    \left(\frac{1}{1}\right)+ \left(\frac{1+2}{1.2}\right)+ \left(\frac{1+2+3}{1.2.3}\right)+ \left(\frac{1+2+3+4}{1.2.3.4}\right)+...................
    find out the \bigcup r. and put it into partial fraction. Through this \sum_{r=1}^{\propto} \bigcup r= \frac{3e}{2} In this case i write the \bigcup{r}
    like this \bigcup r= \frac{\sum_{r=1}^{\propto}}{r!} Then how can i partial fraction.>>>>>>
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  2. #2
    Member srirahulan's Avatar
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    Lightbulb Re: Series

    A little correction \bigcup{r}=\frac{\sum_{r=1}^{\propto}r}{r!}
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  3. #3
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    Re: Series

    I have no clue what you are asking. Please try stating what you are trying to prove again? I am not sure where partial fractions comes into this discussion. Nor do I have any idea what the end result is supposed to be.
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  4. #4
    Member srirahulan's Avatar
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    Exclamation Re: Series

    I ask my \bigcup{r}=\frac{\sum_{r=1}^{\propto}r}{r!} Is correct or not, and then How can i put it into partial fractions.......
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  5. #5
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    Re: Series

    \bigcup r means the union of the sets r. \sum_{r=1}^\infty r is not defined on sets. Then, you are dividing by r!. The notation you are using is meaningless. The only place you define r is in the sum. But then you use r outside the sum, as well. Nowhere do you have anything that would allow partial fractions to be applicable.

    Edit: from the series you tried to write, it looks like you are looking for:

    \sum_{n\ge 1}\dfrac{\sum_{r = 1}^n r}{n!}

    If that is what you mean, then what do you think partial fractions are?
    Last edited by SlipEternal; October 23rd 2013 at 06:07 PM.
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  6. #6
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    Re: Series

    Here is something you can do, but it is not partial fractions:

    \begin{align*}& 1 + \dfrac{1+2}{2!} + \dfrac{1+2+3}{3!} + \cdots \\ = & \sum_{n\ge 1}\dfrac{\binom{n+1}{2}}{n!} \\ = & \sum_{n\ge 1}\dfrac{\left(\dfrac{(n+1)!}{2!(n-1)!}\right)}{n!} \\ = & \sum_{n\ge 1}\dfrac{(n+1)}{2!(n-1)!} \\ = & \sum_{n\ge 1}\dfrac{(n-1)+2}{2!(n-1)!} \\ = & \sum_{n\ge 1}\dfrac{n-1}{2!(n-1)!} + \sum_{n\ge 1}\dfrac{2}{2!(n-1)!} \\ = & \dfrac{1}{2}\sum_{n\ge 0}\dfrac{n}{n!} + \sum_{n\ge 0}\dfrac{1}{n!} \\ = & \dfrac{1}{2}\sum_{n\ge 0}\dfrac{1}{n!} + e \\ = & \dfrac{e}{2} + e \end{align*}

    That gets you what you want.
    Last edited by SlipEternal; October 23rd 2013 at 06:58 PM.
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