# Series Problem and Simplification

1. Mar 28, 2006

### verd

Hi,

I'm a bit confused when it comes to the simplification here. I must've had a brain fart or something when we were going over this, but if I had the following series:

$$\sum _{n=1} ^{\infty} 2^{2n}3^{1-n}$$

I know enough to go this far:
$$\sum _{n=1} ^{\infty} \frac{2^{2n}}{3^{n-1}}$$

But my text book goes on to make the above into this:
$$\sum _{n=1} ^{\infty} 4(\frac{4}{3})^{n-1}$$

I know that this is just a simplification, but how do they get from my second step to the above step? I thought I knew how, but this proved wrong on a couple of different problems I was working on.

Also, for this series:
$$\sum _{n=1} ^{\infty} \frac{3^n+2^n}{6^n}$$

I've done this:
$$\sum _{n=1} ^{\infty} \frac{3^n+2^n}{6^n} = \sum _{n=1} ^{\infty} \frac{3^n}{6^n}+\sum _{n=1} ^{\infty} \frac{2^n}{6^n} = \sum _{n=1} ^{\infty} (\frac{1}{2})^n + \sum _{n=1} ^{\infty} (\frac{1}{3})^n$$

So for the first sum, a=1 and r=1/2, and for the second sum a=1 and r=1/3.

But this is apparently wrong. From the solution manual, for the first one, a=1/2 and r=1/2, for the second one, a=1/3 and r=1/3.

Can someone explain this?

Thanks!

Last edited: Mar 28, 2006
2. Mar 28, 2006

### buzzmath

Use (x^n)^m=x^(mn)

3. Mar 28, 2006

### nocturnal

$$\sum _{n=1}^{\infty} \frac{2^{2n}}{3^{n-1}} = \sum_{n=1}^{\infty} \frac{(2^2)2^{2n-2}}{3^{n-1}} = \sum_{n=1}^{\infty} 4\left( \frac{2^2}{3} \right)^{n-1} = \sum _{n=1} ^{\infty} 4\left( \frac{4}{3} \right)^{n-1}$$

It might help if you expand each sum,

$$\sum _{n=1} ^{\infty} (\frac{1}{2})^n = \frac{1}{2} + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^2 + ...$$

$$\sum _{n=1} ^{\infty} (\frac{1}{3})^n = \frac{1}{3} + \left(\frac{1}{3}\right) \left(\frac{1}{3} \right) + \left(\frac{1}{3} \right)\left(\frac{1}{3} \right)^2 + ...$$

Last edited: Mar 28, 2006
4. Mar 28, 2006

### verd

Thanks for your help. I get the first part with the simplification.

The second part though, I'm having a bit of trouble understanding... I don't understand why my answer wouldn't work.

By that rationale, you'd have:
$$\sum _{n=1} ^{\infty} (\frac{1}{2})^n = \frac{1}{2} + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + ...$$

I mean, it seems that no matter what, it's $$ar^n$$, and if my r is just 1/2, it'd be 1*r^n.... But then the whole a/1-r rule isn't working.

They look like they'd give the same answers. Is there some sort of rule or something that says that you have to do it this way? I'm still a bit confused... ahy.

5. Mar 28, 2006

### nocturnal

compare the expansion to the general equation of a geometric series

$$a + ar + ar^2 + ...$$

pay particularly close attention to the first term of each sum.

Last edited: Mar 28, 2006
6. Mar 28, 2006

### verd

that seems like it'd make sense if n started at 0...

$$\sum _{n=1} ^{\infty} ar^n$$
would give you:

$$ar^1+ar^2+ar^3+ar^4+...$$
wouldn't it?

7. Mar 28, 2006

### verd

ah. I see. It's

$$\sum _{n=1} ^{\infty} ar^{n-1}$$

That changes everything.