series: having trouble rearranging as geom. series

[sphynx_000]

I'm having some trouble setting it into the ar^(n - 1) geometric-series form. Here is the problem:

. . .\(\displaystyle \L \sum_{n=0}^{\infty}\, \frac{4^n}{3^{2n\, +\, 1}}\, =\, \sum_{n=0}^{\infty}\, \frac{4^{n\, -\, 1}}{3^{2n}} \, =\, \sum_{n=0}^{\infty}\, \frac{4^{n\, -\, 1}}{9^n}\, =\, \frac{1}{9}\, \sum_{n=0}^{\infty}\, \left(\frac{4}{9}\right)^{n\, -\, 1}\)


. . .\(\displaystyle \L \mbox{converges to:}\)


. . .\(\displaystyle \L \left(\frac{1}{1\, -\, \frac{4}{9}}\right)\left(\frac{1}{9}\right)\, =\, \begin{array}{|c|}\hline\\\frac{1}{5}?\\\hline\end{array}\)


. . .\(\displaystyle \L \mbox{with }a\,=\,1\,\mbox{ and }\, r\,=\, \frac{4}{9}\)

I was hoping someone could just look over it and make sure it was done correctly. I'm not sure if it was done correctly or not...?

 

[galactus]

YOu have a slight error.

\(\displaystyle \L\\\frac{4^{n}}{3^{2n+1}}\neq\frac{4^{n-1}}{3^{2n}}\)

Actually, \(\displaystyle \L\\\frac{4^{n-1}}{3^{2n}}=\frac{(\frac{4}{9})^{n}}{4}\)

and

\(\displaystyle \L\\\frac{4^{n}}{3^{2n+1}}=\frac{(\frac{4}{9})^{n}}{3}\)

The latter is the one you need.

\(\displaystyle \L\\\frac{1}{3}\sum_{n=0}^{\infty}(\frac{4}{9})^{n}=\frac{1}{3}\left(\frac{1}{1-\frac{4}{9}}\right)=\frac{3}{5}\)

 

[sphynx_000]

could you work out the algebra step by step?

also, im confused weather I should be solving for r^n-1 or r^n....

it depends if I start with 0 to infinity, or 1 to infinity?

 

[pka]

could you work out the algebra step by step?
(Does) it depends if I start with 0 to infinity, or 1 to infinity?

I guess that the question is: "what more do you want?"
Your problem is simply basic basic algebra.
The first reply you got was very complete.
All the steps are shown is great detail.
Now if you cannot follow what was given to you, well I can't think what more there is to do.

As for you second question.
It makes no difference where you start: 1 , 0 , -10 or 1000.
The formula is the same: THE FIRST TERM divided by 1 minus the ratio.

EXAMPLE:\(\displaystyle \L\sum\limits_{k = 10}^\infty {\frac{3}{{5^k }}} = \frac{{\frac{3}{{5^{10} }}}}{{1 - \frac{1}{5}}}\)

 

[sphynx_000]

nevermind,

i found where I was making my error