# Find the sum of the convergent series?

• Jun 1st 2008, 08:14 PM
CrazyLond
Find the sum of the convergent series?
I figured an equation that works for this series but have no idea how to calculate it once I have the sum...none of the rules in our book seem to apply.

http://img48.imageshack.us/img48/7261/9244es7.jpg
• Jun 1st 2008, 08:25 PM
TheEmptySet
Quote:

Originally Posted by CrazyLond
I figured an equation that works for this series but have no idea how to calculate it once I have the sum...none of the rules in our book seem to apply.

http://img48.imageshack.us/img48/7261/9244es7.jpg

Try to use some algebra

$\displaystyle \left( \frac{1}{2}\right)^{2n-5}=\left( \frac{1}{2}\right)^{2(n-1)-3}=\left( \frac{1}{4}\right)^{n-1} \left( \frac{1}{2}\right)^{-3} =8\left( \frac{1}{4}\right)^{n-1}$

So your sum is the same as

$\displaystyle 8\sum_{n=1}^{\infty}\left( \frac{3}{4} \right)^{n-1}$

So $\displaystyle r=\frac{3}{4}$

So the sum of the series $\displaystyle 8\cdot\frac{1}{1-\frac{3}{4}}=8\cdot\frac{1}{\frac{1}{4}}=32$

P.S. I think your index starts at n=1 not at n =0.
• Jun 1st 2008, 09:43 PM
Soroban
Hello, CrazyLond!

Quote:

$\displaystyle S \;=\;8 + 6 + \frac{9}{2} + \frac{27}{8} + \hdots$

Geometric series: first term $\displaystyle a = 8$, common ratio $\displaystyle r = \frac{3}{4}$

Its sum is: .$\displaystyle S \;=\;\frac{8}{1 - \frac{3}{4}} \;=\;32$

• Jun 1st 2008, 10:19 PM
angel.white
Quote:

Originally Posted by CrazyLond
I figured an equation that works for this series but have no idea how to calculate it once I have the sum...none of the rules in our book seem to apply.

http://img48.imageshack.us/img48/7261/9244es7.jpg

I find it easier to look at these like this:

$\displaystyle a_0 = 8$

$\displaystyle a_n = \frac 34a_{n-1}$

Looking at that a little bit (you can do the first few a_n if it helps to see, then you will be familiar enough to spot it right off) it should be clear that $\displaystyle a_n = 8*\left(\frac 34\right)^n$

And since you want to sum all of them, you get

$\displaystyle \sum_{n=0}^\infty a_n$

$\displaystyle =\sum_{n=0}^\infty 8*\left(\frac 34\right)^n$

$\displaystyle =8\sum_{n=0}^\infty \left(\frac 34\right)^n$

At which point you have the same geometric series that TheEmptySet and Soroban solved for.