Sum to Infinity of a Geometric Series

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SUMMARY

The discussion focuses on determining the range of values for x that allow the sum to infinity of two geometric series to exist. For the first series, 1 + 1/x + 1/x² + ..., the condition for convergence is |1/x| < 1, leading to the conclusion that |x| > 1. For the second series, 1/3 + 2x/9 + 4x²/27 + ..., the ratio is derived as r = 6x/9, resulting in the convergence condition -3/2 < x < 3/2. These conclusions align with the textbook answers provided.

PREREQUISITES
  • Understanding of geometric series and their convergence criteria
  • Familiarity with the formula for the sum to infinity, S∞ = a / (1 - r)
  • Basic algebraic manipulation skills
  • Knowledge of absolute values and inequalities
NEXT STEPS
  • Study the properties of geometric series and their convergence
  • Learn about the implications of the ratio test in series convergence
  • Explore advanced topics in series, such as power series and Taylor series
  • Practice solving similar problems involving sums to infinity of geometric series
USEFUL FOR

Students studying calculus, particularly those focusing on series and sequences, as well as educators looking for examples of geometric series convergence criteria.

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Homework Statement



Q. Find the range of values of x for which the sum to infinity exists for each of these series:

(i) 1 + [itex]\frac{1}{x}[/itex] + [itex]\frac{1}{x^2}[/itex] + [itex]\frac{1}{x^3}[/itex] + ...

(ii) [itex]\frac{1}{3}[/itex] + [itex]\frac{2x}{9}[/itex] + [itex]\frac{4x^2}{27}[/itex] + [itex]\frac{8x^3}{81}[/itex] + ...

Homework Equations



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

The Attempt at a Solution



(i) r = [itex]\frac{1}{x}[/itex]/ 1 = [itex]\frac{1}{x}[/itex] [itex]\Rightarrow[/itex] 1 = x
Ans.: From textbook: IxI > 1

(ii) r = [itex]\frac{2x}{9}[/itex]/ [itex]\frac{1}{3}[/itex] = [itex]\frac{6x}{9}[/itex] [itex]\Rightarrow[/itex] 6x = 9 [itex]\Rightarrow[/itex] x = [itex]\frac{9}{6}[/itex] [itex]\Rightarrow[/itex] x = [itex]\frac{3}{2}[/itex]

Ans.: From textbook: -[itex]\frac{3}{2}[/itex] < x < [itex]\frac{3}{2}[/itex]

I'm confused as to whether I'm approaching this correctly, or if I've simply gone wrong in expressing the answers I found. Can someone help me figure this out? Thanks.
 
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For your sum to infinity to exist

Sn must converge as r→∞.

i.e. for |r| < 1

so in your first one, you correctly found r as r = 1/x so it would converge for |1/x| < 1 and you know that |X|< A ⇒ -A<X<A.
 
Ok, I think I see it now. Thanks for clearing that up.
 

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