Sum of infinite geometric series

[motornoob101]

Homework Statement

Homework Equations

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!

 

[Gib Z]

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

 

[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

http://p3t3rl1.googlepages.com/Eqn001.gif

 

[tiny-tim]

Hi motornoob101!

You got confused with your second ∑.

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

 

[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?

 

[HallsofIvy]

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
$$\frac{e^n}{3^{n-1}}= e\left(\frac{e}{3}\right)^{n-1}$$

 

[motornoob101]

Ohh I see. Thanks.