# Telescoping Sum

#### xpack

I have a problem similar to this this thread

http://www.mathhelpforum.com/math-help/analysis-topology-differential-geometry/143798-telescoping-sum-series.html

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

And I know that you use partial fractions and get

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

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

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

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

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{3}{4}$$

What am I doing wrong?

#### Jhevon

MHF Hall of Honor
I have a problem similar to this this thread

http://www.mathhelpforum.com/math-help/analysis-topology-differential-geometry/143798-telescoping-sum-series.html

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

And I know that you use partial fractions and get

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

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

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

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

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{3}{4}$$

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 $$\displaystyle \frac 12 \left( \frac 1n - \frac 1{n + 2} \right)$$. Just a bit easier for the arithmetic, and I got the right answer.

#### xpack

No I messed up with

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

it was suppose to be

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

so now I get

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

which is

$$\displaystyle \frac{1}{4}$$

but if I added them they would be

$$\displaystyle \frac{3}{4}$$

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

#### Jhevon

MHF Hall of Honor
No I messed up with

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

it was suppose to be

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

so now I get

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

which is

$$\displaystyle \frac{1}{4}$$

but if I added them they would be

$$\displaystyle \frac{3}{4}$$

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 $$\displaystyle \frac 1{2n} - \frac 1{2n + 4}$$, 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, $$\displaystyle \lim_{n \to \infty} \left( \frac 12 - \frac 1{2n + 4} \right) \ne \frac 12 - \frac 14$$. That is 1/2. But that term is wrong anyway, that is not the expanded form.