The increment is 1/√(n+1)−2/[√(n+1)+√n]<0, so the sequence decreases. Integral comparison gives Σₖ₌₁ⁿ1/√k≥∫₁ⁿt^(−1/2)dt=2(√n−1), hence x_n≥0; also x_n≤x₁=1. A bounded monotone sequence converges. **B,D**.
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