Sum to Infinity of a Geometric Series

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SUMMARY

The discussion focuses on finding the sum to infinity of a geometric series represented by the formula S = a / (1 - r). The initial term (a) is identified as 1, while the common ratio (r) is calculated as (2x / (x + 1)). The correct sum to infinity is derived as (x + 1) / (1 - 2x), which simplifies to (x + 1) / (1 - x) through algebraic manipulation. Participants clarify the transition from 2x to x in the final expression, ensuring a comprehensive understanding of the geometric series summation.

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  • Understanding of geometric series and their properties
  • Familiarity with algebraic manipulation techniques
  • Knowledge of limits and convergence criteria for series
  • Basic calculus concepts related to infinite series
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  • Study the derivation of the geometric series sum formula S = a / (1 - r)
  • Explore algebraic simplification techniques for rational expressions
  • Learn about convergence tests for infinite series
  • Investigate applications of geometric series in real-world scenarios
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Students studying calculus, mathematics educators, and anyone interested in mastering the concepts of infinite series and geometric progressions.

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Homework Statement



Q. Find, in terms of x, the sum to infinity of the series...

1 + [itex](\frac{2x}{x + 1})[/itex] + [itex](\frac{2x}{x + 1})^2[/itex] + ...

Homework Equations



S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]

The Attempt at a Solution



S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]

a = 1

r = U2/ U1 = [itex](\frac{2x}{x + 1})[/itex]/ 1 = [itex]\frac{2x}{x + 1}[/itex]

[itex]\frac{1}{1 - (2x/ x + 1)}[/itex]

[itex]\frac{x + 1}{1 - 2x}[/itex]

Ans.: From textbook: [itex]\frac{x + 1}{1 - x}[/itex]

Can anyone help me figure out where the 2x becomes just x? Thank you.
 
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If you take

[tex]\frac{1}{x-\frac{2x}{x+1}} \times \frac{x+1}{x+1}[/tex]

you will get

[tex]\frac{x+1}{1(x+1)-2x}[/tex]

Which simplifies to the answer you want.
 
Thanks for clearing that up. I appreciate the help.
 

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