motornoob101
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Sum of infinite geometric series
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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.
motornoob101
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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
http://p3t3rl1.googlepages.com/Eqn001.gif
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Hi motornoob101!
You got confused with your second ∑.
It should be e∑(e/3)^x-1, shouldn't it?
You got confused with your second ∑.
It should be e∑(e/3)^x-1, shouldn't it?
motornoob101
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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?
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You have, skipping over your second term which simply makes no sense,
[tex]\frac{e^n}{3^{n-1}}= \frac{1}{e}\left(\frac{e}{3}\right)^{n-1}[/tex]
which is wrong- you have factored and e out of the numerator: it should be
[tex]\frac{e^n}{3^{n-1}}= e\left(\frac{e}{3}\right)^{n-1}[/tex]
[tex]\frac{e^n}{3^{n-1}}= \frac{1}{e}\left(\frac{e}{3}\right)^{n-1}[/tex]
which is wrong- you have factored and e out of the numerator: it should be
[tex]\frac{e^n}{3^{n-1}}= e\left(\frac{e}{3}\right)^{n-1}[/tex]
motornoob101
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Ohh I see. Thanks.