More telescopic sums

**tsal15** · October 25th 2008, 11:19 AM

My question is attached. Thanks...

**Krizalid** · October 25th 2008, 06:42 PM

$\begin{aligned} \frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ & =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right). \end{aligned}$

Hence, $\sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.$

**Krizalid** · October 25th 2008, 06:56 PM

Here're another solutions with somewhat advanced techniques.

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

$\begin{aligned} \sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}&=\int_{0}^{1}{\int_{0}^{1 }{\left\{ y\sum\limits_{n=1}^{\infty }{\frac{(xy)^{n}}{n}} \right\}\,dx}\,dy} \\ & =-\int_{0}^{1}{\int_{0}^{1}{y\ln (1-xy)\,dx}\,dy} \\ & =\frac{1}{4}. \end{aligned}$

**tsal15** · October 25th 2008, 07:28 PM

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

Originally Posted by Krizalid

$\begin{aligned} \frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ & =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right). \end{aligned}$

Hence, $\sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.$

**tsal15** · October 25th 2008, 08:17 PM

Originally Posted by Krizalid

$\begin{aligned} \frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ & =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right). \end{aligned}$

Hence, $\sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.$

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

**Krizalid** · October 25th 2008, 08:17 PM

Originally Posted by tsal15

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.

Originally Posted by tsal15

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."

**tsal15** · October 27th 2008, 09:08 PM

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 : $n(n+1)(n+2)$

**Jhevon** · October 28th 2008, 07:02 AM

Originally Posted by tsal15

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

**tsal15** · October 28th 2008, 07:51 PM

Originally Posted by Jhevon

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.

**tsal15** · November 1st 2008, 11:30 PM

I'm putting this thread up again so that I can get someone to help me. urgent help needed. Thank you all.

**tsal15** · November 2nd 2008, 05:25 PM

Originally Posted by Krizalid

$\begin{aligned} \frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ & =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right). \end{aligned}$

Hence, $\sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.$

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

**tsal15** · November 2nd 2008, 05:38 PM

Originally Posted by Krizalid

$\begin{aligned} \frac{1}{n(n+1)(n+2)}&=\frac{n+1-n}{n(n+1)(n+2)} \\ & =\frac{n+2-n}{2n(n+2)}-\frac{n+2-(n+1)}{(n+1)(n+2)} \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right) \\ & =\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right)-\left( \frac{1}{n+1}-\frac{1}{n+2} \right). \end{aligned}$

Hence, $\sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.$

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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