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?