Double Summation Result: \alpha^i\alpha^j

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Discussion Overview

The discussion centers around the evaluation of the double summation ##\sum\limits_{i \neq j}^{\infty}\alpha^i\alpha^j##, where ##i, j = 0, 1, 2, ...##. Participants explore methods to express this summation and apply geometric series results to derive a solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant asks for the result of the double summation involving the terms ##\alpha^i\alpha^j##.
  • Another participant suggests expressing the summation as an iterated sum and applying the geometric series result, although they do not provide a specific outcome.
  • A third participant claims to have a final result of ##\frac{2\alpha}{(1+\alpha)(1-\alpha)^2}## but expresses confusion about not arriving at this result using the geometric series method.
  • A fourth participant inquires about the methods used and where the confusion lies, prompting further exploration of the problem.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the evaluation of the summation, with some expressing confusion and others seeking clarification on the methods used.

Contextual Notes

The discussion includes unresolved aspects regarding the application of geometric series and the conditions under which the summation is evaluated.

feryee
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what is the result for the following double summation:
##\sum\limits_{i \neq j}^{\infty}\alpha^i\alpha^j ##

where ## i, j =0,1,2,...##
 
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Express it as an iterated sum and then apply the result for a sum of a geometric series. What do you get?
 
micromass said:
Express it as an iterated sum and then apply the result for a sum of a geometric series. What do you get?
Well actually i have the final result but simply i couldn't get the same answer using geometric sum. Here is the final result:
##\frac{2\alpha}{(1+\alpha)(1-\alpha)^2}##

How is it possible.
 
Have you tried anything? Where are you stuck?
 

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