Telescoping Sum

Sep 2009
22
0
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

\(\displaystyle \frac{1}{2}\)

and the answer is

\(\displaystyle \frac{3}{4}\)

What am I doing wrong?
 

Jhevon

MHF Hall of Honor
Feb 2007
11,663
4,210
New York, USA
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

\(\displaystyle \frac{1}{2}\)

and the answer is

\(\displaystyle \frac{3}{4}\)

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.
 
Sep 2009
22
0
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
Feb 2007
11,663
4,210
New York, USA
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.