Sum of infinite geometric series

motornoob101
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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!
 

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  • Eqn001.gif
    Eqn001.gif
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Try posting the question in LaTex from now on, attachments have to be approved and put extra load on the moderators.
 
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
 
Hi motornoob101!

You got confused with your second ∑.

It should be e∑(e/3)^x-1, shouldn't it? :rolleyes:
 
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?
 
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}
 
Ohh I see. Thanks.
 

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