3 big questions, (1) small questions. ∫n→∞,1/√ (2n+1) ~1√ (2n), ∴ series ∑ [(- 1) n]/√ (2n+65438+)

(3) Let an = [(-1) n] √ [n/(n+1)]. ∵ lim (n →∞) an = (- 1) n ≠ 0。 According to the necessary condition of series convergence, it diverges.

(6) Let an = [(- 1) (n- 1)] n? /2^n。 lim((n→∞)(an+ 1)/an =- 1/2 & lt； 1。 According to the ratio discrimination method, it converges; And lim ((n →∞) è (an+1)/anè =1/2 <1also converges. So the sequence ∑ [(- 1) (n- 1)] n? /2 n convergence, absolute convergence.

2 big questions (2) small questions, ∫ ρ = lim ((n→∞) (an+1)/an =1/2

(5) For small questions, π/2 (n/2) → 0, ∴ sin [π/2 (n/2)] ~ π/2 (n/2). Easy to get, ρ = lim ((n →∞) (an+1)/an = lim ((n →∞) [π/2 (n/2+1/2)]/[π/2 (n/2)] = 60. 1, ∴ According to the ratio discrimination method, the series converges.

For reference.