# Telescoping Series

• Jun 30th 2008, 05:06 PM
auslmar
Telescoping Series
$\displaystyle a_n = \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)}$

It is known that limit of the terms is $\displaystyle 0$ and that the series converges to $\displaystyle \frac{-1}{6}$.

I have a problem that asks for $\displaystyle c_n$ where $\displaystyle a_n = c_n - c_{n+1}$.

Maybe it is simple, but I'm having trouble envisioning how to find $\displaystyle c_n$. If anyone has any tips or hints, I'd greatly appreciate it.

-Austin Martin
• Jun 30th 2008, 05:10 PM
Jhevon
Quote:

Originally Posted by auslmar
$\displaystyle a_n = \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)}$

It is known that limit of the terms is $\displaystyle 0$ and that the series converges to $\displaystyle \frac{-1}{6}$.

I have a problem that asks for $\displaystyle c_n$ where $\displaystyle a_n = c_n - c_{n+1}$.

Maybe it is simple, but I'm having trouble envisioning how to find $\displaystyle c_n$. If anyone has any tips or hints, I'd greatly appreciate it.

-Austin Martin

do you know how to find partial fractions?

$\displaystyle \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)} = - \frac 12 \sum^{\infty}_{n=1} {\color{red} \frac{1}{(n+2)(n+3)}}$

can you find the partial fractions for what is in red? do that, then write out the first few terms, then the last few terms, as many as you need to see a pattern of what would cancel out. then take the limit as $\displaystyle n \to \infty$ of what's left
• Jun 30th 2008, 05:12 PM
auslmar
Quote:

Originally Posted by Jhevon
do you know how to find partial fractions?

$\displaystyle \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)} = - \frac 12 \sum^{\infty}_{n=1} {\color{red} \frac{1}{(n+2)(n+3)}}$

can you find the partial fractions for what is in red? do that, then write out the first few terms, then the last few terms, as many as you need to see a pattern of what would cancel out. then take the limit as $\displaystyle n \to \infty$ of what's left

Oh yes, I've already found the answer to that. I'm just having trouble identifying "$\displaystyle c_n$."
• Jun 30th 2008, 07:46 PM
Mathstud28
Quote:

Originally Posted by auslmar
$\displaystyle a_n = \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)}$

It is known that limit of the terms is $\displaystyle 0$ and that the series converges to $\displaystyle \frac{-1}{6}$.

I have a problem that asks for $\displaystyle c_n$ where $\displaystyle a_n = c_n - c_{n+1}$.

Maybe it is simple, but I'm having trouble envisioning how to find $\displaystyle c_n$. If anyone has any tips or hints, I'd greatly appreciate it.

-Austin Martin

$\displaystyle \frac{1}{(n+2)(n+3)}=\frac{1}{n+2}-\frac{1}{n+3}$

$\displaystyle \therefore\quad{c_n=\frac{1}{n+2}}$

I assume that this is what you are asking
• Jul 1st 2008, 05:00 PM
galactus
Just for kicks, do you want to see the telescope?. Perhaps you already know it, but here is how they look at it and get -1/6 in the event you don't.

Using $\displaystyle \frac{1}{2(n+3)}-\frac{1}{2(n+2)}$

$\displaystyle \left(\frac{1}{8}-\frac{1}{6}\right)+\left(\frac{1}{10}-\frac{1}{8}\right)+\left(\frac{1}{12}-\frac{1}{10}\right)+\left(\frac{1}{14}-\frac{1}{12}\right)+.............$

As you can see, all the terms cancel one another out except the -1/6.

See?.