1. ## Telescoping Sum

I have a problem similar to this this thread

http://www.mathhelpforum.com/math-he...um-series.html

But my problem is
infinity
$
\sum,(\frac{1}{n(n+2)})
$

n=1

And I know that you use partial fractions and get

$1=A(n+2)+B(n)$
or
infinity
$
\sum,(\frac{1}{2n}-\frac{1}{2n+4})
$

n=1

and expanded I have
$\frac{1}{2}-\frac{1}{n+1}$

And I have to find the sum of the series but when I do I get it is equal to

lim $\sum,(\frac{1}{2}-\frac{1}{n+1})$
n->infinity

$\frac{1}{2}$

$\frac{3}{4}$

What am I doing wrong?

2. Originally Posted by xpack
I have a problem similar to this this thread

http://www.mathhelpforum.com/math-he...um-series.html

But my problem is
infinity
$
\sum,(\frac{1}{n(n+2)})
$

n=1

And I know that you use partial fractions and get

$1=A(n+2)+B(n)$
or
infinity
$
\sum,(\frac{1}{2n}-\frac{1}{2n+4})
$

n=1

and expanded I have
$\frac{1}{2}-\frac{1}{n+1}$

And I have to find the sum of the series but when I do I get it is equal to

lim $\sum,(\frac{1}{2}-\frac{1}{n+1})$
n->infinity

$\frac{1}{2}$

$\frac{3}{4}$

What am I doing wrong?
you almost surely canceled too many terms. Check your simplification again. Yup, that's what happened. I did it and it works fine. You partial fraction decomposition is correct and everything. Though I used the form $\frac 12 \left( \frac 1n - \frac 1{n + 2} \right)$. Just a bit easier for the arithmetic, and I got the right answer.

3. No I messed up with

The expanded form
$\frac{1}{2}-\frac{1}{n+1}$

it was suppose to be

$\frac{1}{2}-\frac{1}{2n+4}$

so now I get

$\frac{1}{2}-\frac{1}{4}$

which is

$\frac{1}{4}$

but if I added them they would be

$\frac{3}{4}$

So i dont know why i'm getting a minus though

4. Originally Posted by xpack
No I messed up with

The expanded form
$\frac{1}{2}-\frac{1}{n+1}$

it was suppose to be

$\frac{1}{2}-\frac{1}{2n+4}$

so now I get

$\frac{1}{2}-\frac{1}{4}$

which is

$\frac{1}{4}$

but if I added them they would be

$\frac{3}{4}$

So i dont know why i'm getting a minus though
having read your post the first time, I came away with your partial fraction decomp to be $\frac 1{2n} - \frac 1{2n + 4}$, that is the correct one. when you expand that and cancel terms, you would not end up with what you wrote. And by the way, $\lim_{n \to \infty} \left( \frac 12 - \frac 1{2n + 4} \right) \ne \frac 12 - \frac 14$. That is 1/2. But that term is wrong anyway, that is not the expanded form.