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Math Help - [SOLVED] Find sum of infinite series?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Find sum of infinite series?

    Find the sum of the following infinite series:

    (n = 0 to ∞) (1/n![(k = 0 to n) ((k+1).∫(from 0 to 1) 2^(-(k +1)x) dx]8)
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  2. #2
    Super Member fardeen_gen's Avatar
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    Is it something to do with binomial or A.P/G.P/H.P? I cannot make head or tail still.
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  3. #3
    Moo
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    Just a question : can you possibly learn latex ? ^^
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  4. #4
    Super Member fardeen_gen's Avatar
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    \sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)]8}}

    Thanks for the push Moo! Can anyone help now?


    EDIT: Why is this not working(I want to make the square brackets big):
    Code:
    \sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}
    Last edited by fardeen_gen; April 27th 2009 at 11:32 PM.
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  5. #5
    Moo
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    Quote Originally Posted by fardeen_gen View Post
    \sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)]8}}

    Thanks for the push Moo! Can anyone help now?


    EDIT: Why is this not working(I want to make the square brackets big):
    \sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}
    \sum\limits_{n = 0}^{ \infty } \frac{1}{n!}\left[\sum\limits_{k = 0}^{n} (k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8
    I removed the red brackets

    Anyway, for your problem... :

    We have
    \begin{aligned}<br />
\int_0^1 2^{-(k+1)x} ~dx &=\int_0^1 \exp\left(-(k+1)x\ln(2)\right) ~dx \\<br />
&=\left. -\frac{1}{(k+1)\ln(2)} \cdot \exp\left(-(k+1)x\ln(2)\right)\right|_0^1 \\<br />
&=\left. -\frac{1}{(k+1)\ln(2)} \cdot 2^{-(k+1)x}\right|_0^1 \\<br />
&=-\frac{1}{(k+1)\ln(2)} \cdot (2^{-(k+1)}-1)<br />
\end{aligned}


    \int_0^1 2^{-(k+1)x} ~dx = \frac{1}{(k+1)\ln(2)} \cdot \left(1-\frac{1}{2^{k+1}}\right)


    So
    \begin{aligned} \sum_{k=0}^n (k+1)\int_0^1 2^{-(k+1)x} ~dx<br />
&=\frac{1}{\ln(2)} \sum_{k=0}^n 1-\frac{1}{2^{k+1}} \\<br />
&=\frac{1}{\ln(2)} \left(\sum_{k=0}^n 1 -\sum_{k=0}^n \frac{1}{2^{k+1}}\right) \\<br />
&=\frac{1}{\ln(2)} \left((n+1)-\frac 12 \cdot \frac{1-\frac{1}{2^{n+1}}}{1-\frac 12}\right) \\<br />
&=\frac{1}{\ln(2)} \left((n+1)-\left(1-\frac{1}{2^{n+1}}\right)\right)<br />
\end{aligned}

    \begin{aligned} \sum_{k=0}^n (k+1)\int_0^1 2^{-(k+1)x} ~dx<br />
&=\frac{1}{\ln(2)} \left(n+\frac{1}{2^{n+1}}\right) \end{aligned}


    Now, we have your initial series equal to :
    \begin{aligned}<br />
&=\frac{1}{\ln(2)} \sum_{n=0}^\infty \frac{\left(n+\frac{1}{2^{n+1}}\right)}{n!} \\<br />
&=\frac{1}{\ln(2)} \left( \sum_{n=0}^\infty \frac{n}{n!}+\sum_{n=0}^\infty \frac{(\frac 12)^{n+1}}{n!}\right) \\<br />
&=\frac{1}{\ln(2)} \left( \sum_{n=1}^\infty \frac{n}{n!}+\frac 12 \sum_{n=0}^\infty \frac{(\frac 12)^n}{n!}\right) \end{aligned}
    =\frac{1}{\ln(2)} \left( \sum_{n=1}^\infty \frac{1}{(n-1)!}+\frac 12 \sum_{n=0}^\infty \frac{(\frac 12)^n}{n!}\right)<br />
    =\frac{1}{\ln(2)} \left( \sum_{n=0}^\infty \frac{1}{n!}+\frac 12 \sum_{n=0}^\infty \frac{(\frac 12)^n}{n!}\right)

    Now, recall this series : \sum_{n=0}^\infty \frac{x^n}{n!}=e^x
    For the first one : x=1, for the second one, x=1/2

    And it should be finished.

    HOwever, I don't quite see the interest of calculating such a series :/
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  6. #6
    Super Member fardeen_gen's Avatar
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    Thanks Moo!

    The answer: \frac{2e + \sqrt{2}}{2\ln2}

    HOwever, I don't quite see the interest of calculating such a series :/
    I don't see the interest either. But we got to do what we got to do

    EDIT: My typo
    Answer: \frac{2e + \sqrt{e}}{2\ln2}
    Last edited by fardeen_gen; April 28th 2009 at 10:15 PM.
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  7. #7
    Moo
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    Quote Originally Posted by fardeen_gen View Post
    Thanks Moo!

    The answer: \frac{2e + \sqrt{2}}{2\ln2}



    I don't see the interest either. But we got to do what we got to do
    Hmm I think it's rather \frac{2e+\sqrt{{\color{red}e}}}{2 \ln(2)}, isn't it ?

    You're welcome. And you're right ^^
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