Series convergence representation

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Homework Help Overview

The discussion revolves around the convergence of the series \(\sum_{n=0}^\infty (0.5)^n * e^{-jn}\) and its representation as \(\frac{1}{1-0.5e^{-jn}}\). Participants are exploring the nature of this series and the conditions under which it converges.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the presence of the variable \(n\) in the final representation and whether the series can be classified as a geometric series. There is also discussion about the implications of the exponential term in the series.

Discussion Status

Some participants have offered insights regarding the geometric series and have attempted to clarify the representation of the series. There is an acknowledgment of a potential mistake in the original formulation, and the conversation is moving towards a clearer understanding of the series' structure.

Contextual Notes

There is a mention of the original poster's lack of recent experience with series, which may influence their understanding. Additionally, the discussion includes a correction regarding the final answer's format.

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Homework Statement


[tex]\sum_{n=0}^\infty (0.5)^n * e^{-jn}[/tex]

converges into

[tex]\frac{1}{1-0.5e^{-jn}}[/tex]

Prove the convergence.

Homework Equations



Power series, and perhaps taylor & Macclaurin representation of series.

The Attempt at a Solution



This isn't a homework problem, actually. I just saw this series on the poster and wondered why this is the case (I haven't done series for almost 2 years).

I know for sure that the series has to converge since the [tex]0.5^n[/tex] term approaches 0 as n goes to infinity, but I don't understand how the series written above converges into [tex]\frac{1}{1-0.5e^{-jn}}[/tex]. Can anyone explain?
 
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There shouldn't be an n in your final answer, obviously.
But isn't this just an ordinary geometric series?
[tex]\sum_{n = 0}^\infty x^n = \frac{1}{1 - x}[/tex]
 
That's what I was thinking, except that the series is multiplied by an exponential term (with n). And sorry, there was a mistake - there shouldn't be n in the final answer.
 
Don't get confused over a rewriting of something you already knew :smile:
If I'd write it as
[tex]\sum_{n = 0}^\infty \left( \tfrac12 e^{-j} \right)^n,[/tex]
which is obviously possible since [itex](e^a)^b = e^{ab}[/itex], would you see it's the same?
 
[Hits Head]

Doh, of course. Thanks
 

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