- #1
Sum of infinite geometric series
- Thread starter motornoob101
- Start date
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Homework Help Overview
The discussion revolves around the evaluation of an infinite geometric series, with participants examining their attempts to reconcile their solutions with a provided answer from a solution manual.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- Participants express confusion regarding their calculations and the correct formulation of the series. There are attempts to clarify the notation and structure of the series, with some participants questioning the manipulation of terms and factors.
Discussion Status
Several participants are actively engaging with each other's posts to clarify misunderstandings and correct errors in their reasoning. Some guidance has been offered regarding the proper expression of terms, but no consensus has been reached on the overall solution.
Contextual Notes
There are mentions of formatting issues related to posting in LaTeX, which may affect the clarity of the mathematical expressions being discussed. Additionally, some participants are navigating the constraints of using attachments for their work.
- #2
Gib Z
Homework Helper
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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.
- #3
motornoob101
- 45
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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
- #4
tiny-tim
Science Advisor
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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?
- #5
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?
- #6
HallsofIvy
Science Advisor
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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]
- #7
motornoob101
- 45
- 0
Ohh I see. Thanks.
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