Finding the sum of a convergent Series

[Jake4]

Homework Statement

Sum (n=1->infinity) of 2^(n+2)/3^n

Homework Equations

The Attempt at a Solution

I have literally no idea how to attempt this. We beat into the ground the process of testing convergence and finding that, but not how to find the actual SUM.

any help would be fantastic, Thanks guys!

 

[jhae2.718]

This is a geometric series, $$\sum_{n = 1}^\infty \frac{2^{n+2}}{3^n}$$.

For a geometric series in the form $$\sum_{n = 1}^\infty ar^{n-1}$$, if |r|<1, the sum is given by $$\frac{a}{1-r}$$

How can you make your series look like that?

 

[Jake4]

I have, quite literally, no idea...

 

[jhae2.718]

You need to use properties of exponents so that both the 2 and 3 have a power of n-1. Then, what you pull out will be your a, and the terms raised to n-1 will be r...

 

[Jake4]

What types of properties would allow me to do that?

I'm sorry, I'm finishing up this homework assignment, due in 2 hours... and I have an exam on this today in class. I can test for convergence, I can find taylor and maclauren series, but this is the last part of the homework, I have about a page and a half of examples, and I haven't the faintest idea of how to solve them.

 

[Office_Shredder]

He's saying you need to solve for a here:

$$\frac{2^{n+2}}{3^n} = a \frac{2^{n-1}}{3^{n-1}}$$

 

[jhae2.718]

Recall:
$$\begin{align*}<br /> a^ma^n = a^{m+n} \\<br /> \frac{a^m}{a^n} = a^{m/n} \\<br /> \frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n<br /> \end{align*}$$
etc...

 

[Jake4]

Ahhh! as is the case very often in mathematics, the moment it makes sense, you feel like an idiot...

So simple, thank you so much for bearing with me :)

thanks guys!

 

[Jake4]

just absolutely obliterated my exam.. thanks guys :D