Infinite Series

**vinson24** · April 21st 2010, 05:08 AM

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

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

**Soroban** · April 21st 2010, 05:23 AM

Hello, vinson24!

You missed a fraction in your Partial Fractions . . .

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

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

Here's how I'm doing it:

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

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

**vinson24** · April 21st 2010, 06:06 AM

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

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