Question:

If \( x > 0 \), the series \( x + 2x^2 + 3x^3 + \ldots \) is:

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For series of the form \( \sum n x^n \), use the ratio test and note that such series diverge at \( x = 1 \). Always examine endpoints separately.
Updated On: May 26, 2025
  • convergent if \( x > 1 \) and divergent \( x \geq 1 \)
  • convergent if \( x < 1 \) and divergent \( x \geq 1 \)
  • divergent if \( x > 1 \) and convergent \( x > 1 \)
  • convergent if \( x \leq 1 \) and convergent \( x \geq 1 \)
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The Correct Option is B

Solution and Explanation

We are given the series: \[ x + 2x^2 + 3x^3 + 4x^4 + \cdots = \sum_{n=1}^{\infty} n x^n \] This is a power series of the form: \[ \sum_{n=1}^{\infty} n x^n \] To test for convergence, we apply the Ratio Test: Let \( a_n = n x^n \) Then: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)x^{n+1}}{n x^n} \right| = \left| \frac{n+1}{n} \cdot x \right| = \left(1 + \frac{1}{n}\right) x \] Now take the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = x \] So: - The series converges if \( x < 1 \) - The series diverges if \( x \geq 1 \)
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