For the series ∑ x^n /(n+1), what is its interval of convergence?

• A. (-∞, ∞)
• B. (-1, 1)
• C. [-1, 1)
• D. [-1, 1]

Answer: (C)

This asks about the interval of convergence for a power series. The radius of convergence comes from the ratio of successive terms: with terms having coefficient 1/(n+1), the absolute ratio is |x|·(n+1)/(n+2) → |x| as n → ∞. So the series converges for |x| < 1 (and diverges for |x| > 1).

End behavior at the endpoints matters. When x = 1, the series becomes ∑ 1/(n+1), which is the harmonic series and diverges. When x = -1, the series is ∑ (-1)^n/(n+1), which converges by the alternating series test since 1/(n+1) decreases to 0. This convergence at -1 is conditional (not absolute) because the series of absolute values diverges.

Therefore, the interval of convergence includes -1 but not 1, giving [-1, 1).