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  1. #1
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    sum

    evaluate the series \sum_{k\,=\,2}^{\infty }{\frac{1}{(k+1)(k+2)(k+3)}} as many ways as possible
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Ajreb View Post
    evaluate the series \sum_{k\,=\,2}^{\infty }{\frac{1}{(k+1)(k+2)(k+3)}} as many ways as possible
    \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

    So \iiint\frac{dx}{1-x}\bigg|_{x=1}=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n  +2)(n+3)}

    EDIT: Sorry, now I have time to finish.

    So we know that \begin{aligned}\iiint\frac{dx}{1-x}&=\frac{-1}{2}\ln(1-x^2)(1-x)^2+\frac{1}{4}-\frac{3}{2}x+\frac{3}{4}x^2\\<br />
&=\iiint\sum_{n=0}^{\infty}x^n\\<br />
&=\sum_{n=0}^{\infty}\frac{x^{n+3}}{(n+1)(n+2)(n+3  )}~\forall{x}\in[-1,1]<br />
\end{aligned}

    So

    \begin{aligned}\bigg[\frac{-1}{2}\ln(1-x^2)(1-x)^2+\frac{1}{4}-\frac{3}{2}x+\frac{3}{4}x^2\bigg]\bigg|_{x=1}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+  2)(n+3)}\\<br />
&=\frac{1}{4}<br />
\end{aligned}

    So finally \begin{aligned}\sum_{n=2}^{\infty}\frac{1}{(n+1)(n  +2)(n+3)}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+2)(  n+3)}-\frac{1}{6}-\frac{1}{24}\\<br />
&=\frac{1}{24}\quad\blacksquare<br />
\end{aligned}
    Last edited by Mathstud28; November 21st 2008 at 04:54 PM.
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  3. #3
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    Here's my approach:

    Your series equals \int_{0}^{1}{\int_{0}^{1}{\left( y\sum\limits_{k=2}^{\infty }{\frac{(xy)^{k+1}}{k+1}} \right)\,dx}\,dy}, and after some simple calculations this is -\left( \int_{0}^{1}{\frac{y^{3}+3y^{2}-6y}{6}\,dy}+\int_{0}^{1}{(y-1)\ln (1-y)\,dy} \right). First integral is easy to solve, so for the second one I'll perform a double integration solution, hence the series equals \frac{7}{24}+\int_{0}^{1}{\int_{0}^{y}{\frac{y-1}{1-z}\,dz}\,dy}=\frac{7}{24}-\frac{1}{4}=\frac{1}{24}.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Ill just go ahead and be the one to suggest the ugliness that is PFD followed by Telescoping Series for this.

    \sum_{n=2}^{\infty}\left\{\frac{1}{2(n+3)}-\frac{1}{n+2}+\frac{1}{2(n+1)}\right\}

    If anyone would like to do it I would be interested to see how many terms you would have to go through to see the actual sum show its face.
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  5. #5
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    both methods given by Mathstud28 and Krizalid are good but they're not strong enough to handle the general form, i.e. evaluating: \sigma_n=\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)}, where n \geq 2 is given.

    note that the sum here starts from k = 0, which is more usual than k = 2. evaluating \sigma_n is quite easy if you're familiar with gamma and beta functions. here's the solution: first observe that:

    \frac{1}{(k+1)(k+2) \cdots (k+n)}=\frac{k!(n-1)!}{(n-1)!(k+n)!}=\frac{\Gamma(k+1)\Gamma(n)}{(n-1)!\Gamma(k+n+1)}=\frac{B(k+1,n)}{(n-1)!}. from here things are straightforward:

    (n-1)!\sigma_n=\sum_{k=0}^{\infty}B(k+1,n)=\sum_{k=0}  ^{\infty}\int_0^1t^k(1-t)^{n-1} \ dt=\int_0^1[(1-t)^{n-1}\sum_{k=0}^{\infty}t^k] \ dt

    =\int_0^1\frac{(1-t)^{n-1}}{1-t} \ dt = \int_0^1 (1-t)^{n-2} \ dt = \frac{1}{n-1} \Longrightarrow \sigma_n=\frac{1}{(n-1)(n-1)!}. \ \ \Box


    as a result, if you really want your sum to start from k = 2, you will get: \sum_{k=2}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)}=\sigma_n - \frac{1}{n!} - \frac{1}{(n+1)!}=\frac{2}{(n-1)(n+1)!}.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    both methods given by Mathstud28 and Krizalid are good but they're not strong enough to handle the general form, i.e. evaluating: \sigma_n=\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)}, where n \geq 2 is given.

    note that the sum here starts from k = 0, which is more usual than k = 2. evaluating \sigma_n is quite easy if you're familiar with gamma and beta functions. here's the solution: first observe that:

    \frac{1}{(k+1)(k+2) \cdots (k+n)}=\frac{k!(n-1)!}{(n-1)!(k+n)!}=\frac{\Gamma(k+1)\Gamma(n)}{(n-1)!\Gamma(k+n+1)}=\frac{B(k+1,n)}{(n-1)!}. from here things are straightforward:

    (n-1)!\sigma_n=\sum_{k=0}^{\infty}B(k+1,n)=\sum_{k=0}  ^{\infty}\int_0^1t^k(1-t)^{n-1} \ dt=\int_0^1[(1-t)^{n-1}\sum_{k=0}^{\infty}t^k] \ dt

    =\int_0^1\frac{(1-t)^{n-1}}{1-t} \ dt = \int_0^1 (1-t)^{n-2} \ dt = \frac{1}{n-1} \Longrightarrow \sigma_n=\frac{1}{(n-1)(n-1)!}. \ \ \Box


    as a result, if you really want your sum to start from k = 2, you will get: \sum_{k=2}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)}=\sigma_n - \frac{1}{n!} - \frac{1}{(n+1)!}=\frac{2}{(n-1)(n+1)!}.
    Hmm, this is cool because melding my method and yours together

    \underbrace{\int_0^1\int\cdots\int\frac{dx}{1-x}}_{n\text{ times}}=\frac{1}{(n-1)(n-1)!}
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Hmm, this is cool because melding my method and yours together

    \underbrace{\int_0^1\int\cdots\int\frac{dx}{1-x}}_{n\text{ times}}=\frac{1}{(n-1)(n-1)!}
    right! you can rewrite it more clearly: \int_0^1\int_0^{x_n} \cdots \int_0^{x_3} \int_0^{x_2} \frac{\ dx_1 \ dx_2 \ \cdots \ dx_n}{1-x_1} =\frac{1}{(n-1)(n-1)!}, \ \ 0 < x_j < 1.

    the challenge now is to prove this result directly, i.e. without using the series!
    Last edited by NonCommAlg; November 21st 2008 at 11:06 PM.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    right! you can rewrite it more clearly: \underbrace{\int_0^1\int_0^{x_n} \cdots \int_0^{x_3} \int_0^{x_2}}_{n\text{ times}} \frac{\ dx_1 \ dx_2 \ \cdots \ dx_n}{1-x_1} =\frac{1}{(n-1)(n-1)!}, \ \ 0 < x_j < 1.

    the challenge now is to prove this result directly, i.e. without using the series!
    I will come back tomorrow and give it a go, but right now I am thinking induction. And maybe playing around with some mroe generalized functions so that I can use Leibniz's Rule of Differentiation Under the Integral Sign in the inductive step.
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