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: Understanding the Concept:
The given series is an infinite geometric series. The convergence or divergence of such a series depends entirely on the value of its common ratio, \(r\).

Step 2: Key Formula or Approach:
An infinite geometric series \(\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \dots\)

Converges to the sum \(S = \frac{a}{1-r}\) if the absolute value of the common ratio \(|r|<1\).
Diverges if \(|r| \geq 1\). \end{itemize}
Step 3: Detailed Explanation:
The given series is \(1+r+r^2+r^3+\dots\).
The first term is \(a = 1\), and the common ratio is \(r\).
We are also given the condition that \(r>0\).
Applying the convergence rule for a geometric series, the series converges if \(|r|<1\). Since \(r>0\), this simplifies to \(0<r<1\).
The series diverges if \(|r| \geq 1\). Since \(r>0\), this simplifies to \(r \geq 1\).
Therefore, the statement "It converges, if \(0<r<1\) and diverges, if \(r \geq 1\)" is the correct description of the series' behavior.

Step 4: Final Answer:
The statement in option (B) correctly describes the conditions for convergence and divergence of the given geometric series.