1. ## Infinite Series

Show :$\displaystyle (\frac{1}{1*3})+ (\frac{1}{2*4})+ (\frac{1}{3*5})+.....=\frac{3}{4}$

When i do the problem i keep getting $\displaystyle \frac{3}{2}$instead of [tex]\frac{3}{4}/MATH]
Heres how im doing it
$\displaystyle (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+...$
which gives me as the lim n appraoches infinity $\displaystyle (\frac{3}{2}-\frac{1}{n+2})$
I dont see where i am going wrong or how to get $\displaystyle \frac{3}{4}$

2. Hello, vinson24!

You missed a fraction in your Partial Fractions . . .

Show that: .$\displaystyle \frac{1}{1\cdot3}+ \frac{1}{2\cdot4}+ \frac{1}{3\cdot5}+ \hdots \;=\;\frac{3}{4}$

When i do the problem i keep getting $\displaystyle \frac{3}{2}$ instead of $\displaystyle \frac{3}{4}$

Here's how I'm doing it:

$\displaystyle \left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+ \hdots$ . Here!

. . $\displaystyle \frac{1}{n\cdot(n+2)}\;=\;{\color{red}\frac{1}{2}} \,\bigg[\frac{1}{n} - \frac{1}{n+2}\bigg]$

3. With the way you do it I'm getting 1/2