# Sum of infinite geometric series

1. Apr 10, 2008

### motornoob101

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I don't get what I am doing wrong here, I have attached my solution below. The solution manual have their answer as 3e/(3-e). Thanks!

#### Attached Files:

• ###### Eqn001.gif
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2. Apr 10, 2008

### Gib Z

Try posting the question in LaTex from now on, attachments have to be approved and put extra load on the moderators.

3. Apr 10, 2008

### motornoob101

I type really slow in Latex, I type faster in Mathtype in word, that's why I included as an attachment. Sorry. Anyhow, here is the link to that attachment

4. Apr 11, 2008

### tiny-tim

Hi motornoob101!

You got confused with your second ∑.

It should be e∑(e/3)^x-1, shouldn't it?

5. Apr 11, 2008

### motornoob101

Erm I don't think get it.. So I have e^(x-1) which is same as (e^x) (e^-1) and e^-1 is (1/e) no?

6. Apr 11, 2008

### HallsofIvy

Staff Emeritus
You have, skipping over your second term which simply makes no sense,
$$\frac{e^n}{3^{n-1}}= \frac{1}{e}\left(\frac{e}{3}\right)^{n-1}$$
which is wrong- you have factored and e out of the numerator: it should be
$$\frac{e^n}{3^{n-1}}= e\left(\frac{e}{3}\right)^{n-1}$$

7. Apr 11, 2008

### motornoob101

Ohh I see. Thanks.