# Rearranging Equation to determine conv/div

• Mar 30th 2008, 05:00 PM
trev-gdr
Rearranging Equation to determine conv/div
I just found it simpler to take a screenshot. I have no idea how to go about rearranging the equation like the solution does below.

Thanks.
• Mar 30th 2008, 05:23 PM
ThePerfectHacker
Note that,
$\frac{\pi ^n}{3^{n+1}} = \frac{\pi^n}{3^n \cdot 3} = \frac{1}{3} \cdot \frac{\pi^n}{3^n} = \frac{1}{3} \cdot \left( \frac{\pi}{3} \right)^n$
Thus,
$\sum_{n=0}^{\infty}\frac{\pi^n}{3^{n+1}} = \sum_{n=0}^{\infty} \frac{1}{3}\cdot \left( \frac{\pi}{3} \right)^n = \frac{1}{3}\sum_{n=0}^{\infty} \left( \frac{\pi}{3} \right)^n$
• Mar 30th 2008, 05:26 PM
trev-gdr
Ahh, so there is a property that allows 3^(n+1)=3^n + 3^1

Am I right on that?
• Mar 30th 2008, 05:31 PM
ThePerfectHacker
Quote:

Originally Posted by trev-gdr
Ahh, so there is a property that allows 3^(n+1)=3^n + 3^1

Am I right on that?

$3^{n+1} = \underbrace{3\cdot 3\cdot ... \cdot 3}_{n+1 \text{ times }} = \underbrace{3\cdot 3\cdot ... \cdot 3}_{n\text{ times }}\cdot 3 = 3^n \cdot 3$.

In general, $x^{n+m} = x^n \cdot x^m$ where $n,m$ are positive integers.

You supposed to have known this in algebra. Not when learning Calculus.
• Mar 30th 2008, 05:41 PM
trev-gdr
Sorry I put + and thought *.

And it does so happen that people tend to forget things when they haven't done it in a while....hmm what an odd theory :).

Thanks!