Finding the sum of an infinite series

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Homework Help Overview

The discussion revolves around finding the sum of an infinite series involving complex numbers, specifically the series \(\sum _{n=1}^{\infty } \left( i /2\right) ^{2\,n}\). Participants are exploring methods to approach this problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to seek hints for starting the problem. Some participants suggest considering the simplification of \((i/2)^2\) and the use of the geometric series formula. Others propose breaking down \(i\) into its polar form and utilizing Euler's identity, while one participant emphasizes that the series can be treated as a real geometric series.

Discussion Status

The discussion is active, with various approaches being explored. Participants are providing hints and suggestions without reaching a consensus on a single method. There is a focus on different interpretations of the series and its components.

Contextual Notes

Participants are navigating the complexities of working with complex numbers and geometric series, with some expressing uncertainty about the implications of using \(i\) in the series.

szklany
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Find the sum of the infinite series

[tex]\sum _{n=1}^{\infty } \left( i /2\right) ^{2\,n}[/tex]

I just can't seem to get started on this problem, so I was hoping somebody could give me a hint, as to what methods i should read up on.
 
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A good place to start would be to consider what (i/2)^2 is, that simplifies the sum considerably. You also need the formula for geometric series, that you can find for example from wikipedia.
 
Actually, break i into its polar form which is [tex]i^2^n = e^i^n^(^p^i^)[/tex] . Now expand using Euler's identity and you are left only with [tex]cos\ n(pi) = (-1)^n[/tex] From there it is a geometric series.
 
Actually, there is no need to worry about "i". [itex](i/2)^{2n}= (i^2/4)^n= (-1/4)^n[/itex] so this is a purely real geometric series.
 

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