I have a problem similar to this this thread

http://www.mathhelpforum.com/math-he...um-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?