Question

Which of the following statement is true about the geometric series \(1+r+r^2+r^3+.............; (r>0)\) ?

• A. It diverges, if \(0<r<1\) and converges, if \(r \geq 1\)
• B. It converges, if \(0<r<1\) and diverges, if \(r \geq 1\)
• C. It is always convergent
• D. It is always divergent

Hint:

For any geometric series, the absolute value of the common ratio is the key. If \(|r|<1\), it converges. If \(|r| \geq 1\), it diverges. Always remember this fundamental rule.

Answer 1

The Correct Option is B

Step 1: Concept Overview:
The provided series is an infinite geometric series. Its convergence hinges on the common ratio, \(r\).
Step 2: Core Formula:
Consider an infinite geometric series \(\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \dots\).
Convergence: The series converges to \(S = \frac{a}{1-r}\) if \(|r| < 1\).
Divergence: The series diverges if \(|r| \geq 1\).
Step 3: Detailed Analysis:
We have the series \(1 + r + r^2 + r^3 + \dots\).
The initial term is \(a = 1\), and the common ratio is \(r\).
The condition \(r > 0\) is given.
For convergence, the rule \(|r| < 1\) applies. Given \(r > 0\), this becomes \(0 < r < 1\).
For divergence, the rule \(|r| \geq 1\) applies. Given \(r > 0\), this becomes \(r \geq 1\).
Thus, the statement "It converges, if \(0 < r < 1\) and diverges, if \(r \geq 1\)" correctly describes the series' behavior.
Step 4: Conclusion:
Option (B) accurately reflects the convergence and divergence conditions for this geometric series.