# More telescopic sums

• Oct 25th 2008, 12:19 PM
tsal15
More telescopic sums
My question is attached. Thanks...
• Oct 25th 2008, 07:42 PM
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, $\sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.$
• Oct 25th 2008, 07:56 PM
Krizalid
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}
• Oct 25th 2008, 08:28 PM
tsal15
Sorry Krizalid, but could possibly elaborate a little on why your initial equation was $\frac{1}{n(n+1)(n+2)}$ and not $\frac{2}{n(n+1)(n+2)}$? 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
\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, $\sum\limits_{n=1}^{\infty }{\frac{1}{n(n+1)(n+2)}}=\frac{1}{2}\left( 1+\frac{1}{2} \right)-\frac{1}{2}=\frac{1}{4}.$

• Oct 25th 2008, 09:17 PM
tsal15
Quote:

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
• Oct 25th 2008, 09:17 PM
Krizalid
Quote:

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.

Quote:

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."
• Oct 27th 2008, 10:08 PM
tsal15
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 : $n(n+1)(n+2)$
(Crying)
• Oct 28th 2008, 08:02 AM
Jhevon
Quote:

Originally Posted by tsal15
Can anyone else shed some light on this predicament?

i'll try

Quote:

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"

Quote:

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

Quote:

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 : $n(n+1)(n+2)$
(Crying)
please tells us what "anti-difference" means
• Oct 28th 2008, 08:51 PM
tsal15
Quote:

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 (Speechless).

$\sum\limits_{k=15}^{84} \frac{2k+1}{k^2(k+1)^2}$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,

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

• Nov 2nd 2008, 12:30 AM
tsal15
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, 06:25 PM
tsal15
Quote:

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
• Nov 2nd 2008, 06:38 PM
tsal15
Quote:

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:

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