My question is attached. Thanks...

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- Oct 25th 2008, 11:19 AMtsal15More telescopic sums
My question is attached. Thanks...

- Oct 25th 2008, 06:42 PMKrizalid
$\displaystyle \begin{aligned}

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

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

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

& =\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).

\end{aligned}$

Hence, $\displaystyle \sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.$ - Oct 25th 2008, 06:56 PMKrizalid
Here're another solutions with somewhat advanced techniques.

$\displaystyle \begin{aligned}

\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)}} \\

& =\frac{1}{2}\int_{0}^{1}{\left\{ (1-x)^{2}\sum_{n=1}^{\infty }{x^{n-1}} \right\}\,dx} \\

& =\frac{1}{2}\int_{0}^{1}{(1-x)\,dx} \\

& =\frac{1}{4}.

\end{aligned}$

$\displaystyle \begin{aligned}

\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} \\

& =-\int_{0}^{1}{\int_{0}^{1}{y\ln (1-xy)\,dx}\,dy} \\

& =\frac{1}{4}.

\end{aligned}$ - Oct 25th 2008, 07:28 PMtsal15
Sorry Krizalid, but could possibly elaborate a little on why your initial equation was $\displaystyle \frac{1}{n(n+1)(n+2)}$ and not $\displaystyle \frac{2}{n(n+1)(n+2)}$? and also could you show me where in your solution did you use $\displaystyle \frac{-1}{n(n+1)}$ as the antidifference.

Thanks heaps

- Oct 25th 2008, 08:17 PMtsal15
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:

$\displaystyle S_3 = 18 + 36 + ... + 3(n+1)(n+2) + ... + 30906$ using $\displaystyle {{n(n+1)(n+2)}}$ as the antidifference sequence - Oct 25th 2008, 08:17 PMKrizalid
- Oct 27th 2008, 09:08 PMtsal15
Can anyone else shed some light on this predicament?

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

sum $\displaystyle S_1 = \frac{2}{1.2.3}+ \frac{2}{2.3.4} + ... + \frac{2}{n(n+1)(n+2)}+ \frac{2}{100.101.102}$

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

$\displaystyle S_2 = 3.2.3 + 3.3.4 + ... + 3(n+1)(n+2) + ... + 3.101.102$

antidifference : $\displaystyle n(n+1)(n+2)$

(Crying) - Oct 28th 2008, 07:02 AMJhevon
i'll try

Quote:

Where did Krizalid use $\displaystyle \frac{-1}{n(n+1)}$ as the antidifference?

Quote:

also does the whole solution change much if Krizalid was to evaluate $\displaystyle \frac{2}{n(n+1)(n+2)}$ instead of $\displaystyle \frac{1}{n(n+1)(n+2)}$

Quote:

And if Krizalid hasn't used the antidifference, could someone show me how to use the antidifference $\displaystyle \frac{-1}{n(n+1)}$ whilst using telescopic sums method to evaluate the finite

sum $\displaystyle S_1 = \frac{2}{1.2.3}+ \frac{2}{2.3.4} + ... + \frac{2}{n(n+1)(n+2)}+ \frac{2}{100.101.102}$

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

$\displaystyle S_2 = 3.2.3 + 3.3.4 + ... + 3(n+1)(n+2) + ... + 3.101.102$

antidifference : $\displaystyle n(n+1)(n+2)$

(Crying)

- Oct 28th 2008, 07:51 PMtsal15

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 (Speechless).

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

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

Tell me if you want more information. - Nov 1st 2008, 11:30 PMtsal15
I'm putting this thread up again so that I can get someone to help me. urgent help needed. Thank you all.

- Nov 2nd 2008, 05:25 PMtsal15
- Nov 2nd 2008, 05:38 PMtsal15