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Math Help - More telescopic sums

  1. #1
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    More telescopic sums

    My question is attached. Thanks...
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  2. #2
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    \begin{aligned}<br />
   \frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ <br />
 & =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ <br />
 & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ <br />
 & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right).<br />
\end{aligned}

    Hence, \sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.
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  3. #3
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    Here're another solutions with somewhat advanced techniques.

    \begin{aligned}<br />
   \sum_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}&=\frac{1}{2}\sum_{n=1}^{\  infty }{\frac{\Gamma (n)\Gamma (3)}{\Gamma (n+3)}} \\ <br />
 & =\frac{1}{2}\int_{0}^{1}{\left\{ (1-x)^{2}\sum_{n=1}^{\infty }{x^{n-1}} \right\}\,dx} \\ <br />
 & =\frac{1}{2}\int_{0}^{1}{(1-x)\,dx} \\ <br />
 & =\frac{1}{4}.<br />
\end{aligned}

    \begin{aligned}<br />
   \sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}&=\int_{0}^{1}{\int_{0}^{1  }{\left\{ y\sum\limits_{n=1}^{\infty }{\frac{(xy)^{n}}{n}} \right\}\,dx}\,dy} \\ <br />
 & =-\int_{0}^{1}{\int_{0}^{1}{y\ln (1-xy)\,dx}\,dy} \\ <br />
 & =\frac{1}{4}.<br />
\end{aligned}
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  4. #4
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    Sorry Krizalid, but could possibly elaborate a little on why your initial equation was \frac{1}{n(n+1)(n+2)} and not \frac{2}{n(n+1)(n+2)}? and also could you show me where in your solution did you use \frac{-1}{n(n+1)} as the antidifference.

    Thanks heaps

    Quote Originally Posted by Krizalid View Post
    \begin{aligned}<br />
\frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ <br />
& =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ <br />
& =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ <br />
& =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right).<br />
\end{aligned}

    Hence, \sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    \begin{aligned}<br />
\frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ <br />
& =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ <br />
& =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ <br />
& =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right).<br />
\end{aligned}

    Hence, \sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.
    I've also got another telescopic sum question that, if you don't mind, hopefully u can help me with.

    Again i need to use the telescopic sums method to evaluate the following finite sum:

    S_3 = 18 + 36 + ... + 3(n+1)(n+2) + ... + 30906 using {{n(n+1)(n+2)}} as the antidifference sequence
    Last edited by tsal15; November 1st 2008 at 11:28 PM.
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  6. #6
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    Quote Originally Posted by tsal15 View Post

    Sorry Krizalid, but could possibly elaborate a little on why your initial equation was \frac{1}{n(n+1)(n+2)} and not \frac{2}{n(n+1)(n+2)}?
    I think you could tell me what's the big difference.

    Quote Originally Posted by tsal15 View Post

    and also could you show me where in your solution did you use \frac{-1}{n(n+1)} as the antidifference.

    Thanks heaps
    Dunno what's the "antidifference."
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  7. #7
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    Can anyone else shed some light on this predicament?

    Where did Krizalid use \frac{-1}{n(n+1)} as the antidifference? also does the whole solution change much if Krizalid was to evaluate \frac{2}{n(n+1)(n+2)} instead of \frac{1}{n(n+1)(n+2)} And if Krizalid hasn't used the antidifference, could someone show me how to use the antidifference \frac{-1}{n(n+1)} whilst using telescopic sums method to evaluate the finite
    sum S_1 = \frac{2}{1.2.3}+ \frac{2}{2.3.4} + ... + \frac{2}{n(n+1)(n+2)}+ \frac{2}{100.101.102}

    Also using the method of telescopic sums, can someone evaluate the following finite sum using the provide antidifference?

    S_2 = 3.2.3 + 3.3.4 + ... + 3(n+1)(n+2) + ... + 3.101.102
    antidifference : n(n+1)(n+2)
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tsal15 View Post
    Can anyone else shed some light on this predicament?
    i'll try

    Where did Krizalid use \frac{-1}{n(n+1)} as the antidifference?
    i'm with Kriz on this. i don't know what you mean by "anti-difference"

    also does the whole solution change much if Krizalid was to evaluate \frac{2}{n(n+1)(n+2)} instead of \frac{1}{n(n+1)(n+2)}
    no. you would just factor out the 2 and put it in front of the summation sign. then everything follows in the same way, and you multiply by 2 when done

    And if Krizalid hasn't used the antidifference, could someone show me how to use the antidifference \frac{-1}{n(n+1)} whilst using telescopic sums method to evaluate the finite
    sum S_1 = \frac{2}{1.2.3}+ \frac{2}{2.3.4} + ... + \frac{2}{n(n+1)(n+2)}+ \frac{2}{100.101.102}

    Also using the method of telescopic sums, can someone evaluate the following finite sum using the provide antidifference?

    S_2 = 3.2.3 + 3.3.4 + ... + 3(n+1)(n+2) + ... + 3.101.102
    antidifference : n(n+1)(n+2)
    please tells us what "anti-difference" means
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    i'll try

    i'm with Kriz on this. i don't know what you mean by "anti-difference"

    no. you would just factor out the 2 and put it in front of the summation sign. then everything follows in the same way, and you multiply by 2 when done

    please tells us what "anti-difference" means

    Well, I'll give an example which might make it easier. My lecturer is bad at teaching it to me, so hopefully it doesn't pass on .

    \sum\limits_{k=15}^{84} \frac{2k+1}{k^2(k+1)^2}is found, using an antidifference expression for its summand. By partial fractions: \frac{2k+1}{k^2(k+1)^2} = \frac{1}{k^2} - \frac{1}{(k+1)^2} = \frac{-1}{(k+1)^2} - \frac{-1}{k^2} The 2nd expression is a difference exactly as needed. So summand sequence { x_k} = { \frac{2k+1}{k^2(k+1)^2}} has antidifference { T_k} = { \frac{-1}{k^2}}. So,

    \sum\limits_{k=15}^{84} \frac{2k+1}{k^2(k+1)^2} = \frac{-1}{(85)^2} - \frac{-1}{(15)^2} = \frac{56}{13005}

    Tell me if you want more information.
    Last edited by tsal15; October 28th 2008 at 07:54 PM. Reason: latex mistake
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  10. #10
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    Unhappy

    I'm putting this thread up again so that I can get someone to help me. urgent help needed. Thank you all.
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  11. #11
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    Quote Originally Posted by Krizalid View Post
    \begin{aligned}<br />
\frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ <br />
& =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ <br />
& =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ <br />
& =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right).<br />
\end{aligned}

    Hence, \sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.
    Is this also possible?

    \sum\limits_{n=1}^{100} \frac{1}{n(n+1)(n+2)}

    \sum\limits_{n=1}^{100} \frac{1}{n(n+1)(n+3)} - \frac{1}{n(n+1)}

    Also i think you have assumed the the sum is infinite, its actually finite
    Last edited by tsal15; November 2nd 2008 at 06:10 PM.
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  12. #12
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    Quote Originally Posted by Krizalid View Post
    \begin{aligned}<br />
\frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ <br />
& =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ <br />
& =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ <br />
& =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right).<br />
\end{aligned}

    Hence, \sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.
    I've done my calculations over and over, but i can't see how the following are equal:

    \frac{1}{n(n+1)(n+2)} = \frac{n+2-n}{2n(n+2)} - \frac{n+2-(n+1)}{(n+1)(n+2)}

    could u please explain
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