Question
\( x \) values where it converges \( 1+(2 x+1)+(2 x+1)^{2}+\ldots \)
Ask by Klein Montgomery. in South Africa
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The series converges for all real numbers \( x \) between –1 and 0.
Solution
The series you've provided is:
\[
1 + (2x + 1) + (2x + 1)^2 + (2x + 1)^3 + \dots
\]
This is a **geometric series** where the first term \( a = 1 \) and the common ratio \( r = 2x + 1 \).
### Convergence of a Geometric Series
A geometric series converges **if and only if** the absolute value of the common ratio is **less than 1**:
\[
|r| < 1
\]
Substituting \( r = 2x + 1 \):
\[
|2x + 1| < 1
\]
### Solving for \( x \)
Let's solve the inequality:
\[
|2x + 1| < 1
\]
This absolute value inequality translates to:
\[
-1 < 2x + 1 < 1
\]
Subtract 1 from all parts of the inequality:
\[
-2 < 2x < 0
\]
Divide all parts by 2:
\[
-1 < x < 0
\]
### Conclusion
The series **converges** for all real numbers \( x \) in the open interval:
\[
-1 < x < 0
\]
**Therefore, the series converges for all real numbers x between –1 and 0.**
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution The Deep Dive
This series is a geometric series where the first term \( a = 1 \) and the common ratio \( r = (2x + 1) \). For a geometric series to converge, the absolute value of the common ratio must be less than one, so we need \( |2x + 1| < 1 \). This can be solved to find the \( x \) values where the series converges. Solving this inequality gives us \( -1 < 2x + 1 < 1 \). Simplifying leads to the range \( -1 < 2x < 0 \), or \( -\frac{1}{2} < x < 0 \). Thus, the series converges for \( x \) in the interval \((-0.5, 0)\).
