1. ## Telescoping Series

$a_n = \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)}$

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

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

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

-Austin Martin

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

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

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

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

-Austin Martin
do you know how to find partial fractions?

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

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 $n \to \infty$ of what's left

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

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

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 $n \to \infty$ of what's left
Oh yes, I've already found the answer to that. I'm just having trouble identifying " $c_n$."

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

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

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

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

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

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

I assume that this is what you are asking

5. 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 $\frac{1}{2(n+3)}-\frac{1}{2(n+2)}$

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

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

See?.
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