例2判别数列an=(-1)2+ 的收敛性 n+1 解因为 lim(-1) 不存在 lim 0 n→∞n+ 所以 发散
~ 2 Oê αn = (−1)n + i n + 1 Âñ5. ) ÏǑ limn→∞ (−1)n , Ø3, limn→∞ 1 n + 1 = 0. ¤± αn = (−1)n + i n + 1 uÑ. 6/54
例3证明 a>1, lim a= , a<1, 1, 不存在,a=1,a≠1
~ 3 y²: limn→∞ α n = 0, |α| > 1, ∞, |α| < 1, 1, α = 1, Ø3, |α| = 1, α 6= 1. ) α = r(cosθ + isinθ). u´ (1) |α| < 1 , k r < 1, K 0 6 |α n | = |r n (cosnθ + sinnθ)| = r n ÏǑ limn→∞ r n = 0, ¤± limn→∞ α n = 0. (2) |α| > 1 , k 1 α < 1, Kâ (1) (Ø, k limn→∞ 1 α n = 0 ¤± limn→∞ α n = ∞. 7/54
例3证明 a>1, lim a= , a<1, 1, 1, 不存在,a=1,a≠1 解设a=r(cos+isin).于是
~ 3 y²: limn→∞ α n = 0, |α| > 1, ∞, |α| < 1, 1, α = 1, Ø3, |α| = 1, α 6= 1. ) α = r(cosθ + isinθ). u´ (1) |α| < 1 , k r < 1, K 0 6 |α n | = |r n (cosnθ + sinnθ)| = r n ÏǑ limn→∞ r n = 0, ¤± limn→∞ α n = 0. (2) |α| > 1 , k 1 α < 1, Kâ (1) (Ø, k limn→∞ 1 α n = 0 ¤± limn→∞ α n = ∞. 7/54
例3证明 a>1, lim a= , a<1, 1 不存在,a=1,a≠1 解设a=r(cos+isin).于是 (1)当|a|<1时,有r<1,则 0<a"=r"(cosn0+sinne)l=r
~ 3 y²: limn→∞ α n = 0, |α| > 1, ∞, |α| < 1, 1, α = 1, Ø3, |α| = 1, α 6= 1. ) α = r(cosθ + isinθ). u´ (1) |α| < 1 , k r < 1, K 0 6 |α n | = |r n (cosnθ + sinnθ)| = r n ÏǑ limn→∞ r n = 0, ¤± limn→∞ α n = 0. (2) |α| > 1 , k 1 α < 1, Kâ (1) (Ø, k limn→∞ 1 α n = 0 ¤± limn→∞ α n = ∞. 7/54
例3证明 a>1, lim a= , a<1, 1, 不存在,a=1,a≠1 解设a=r(cos+isin).于是 (1)当|a<1时,有r<1,则 0<a"=r"(cosn0+sinne)l=r" 因为 lim rn=0,所以 lim a=o 时根据(1)的结论,有
~ 3 y²: limn→∞ α n = 0, |α| > 1, ∞, |α| < 1, 1, α = 1, Ø3, |α| = 1, α 6= 1. ) α = r(cosθ + isinθ). u´ (1) |α| < 1 , k r < 1, K 0 6 |α n | = |r n (cosnθ + sinnθ)| = r n ÏǑ limn→∞ r n = 0, ¤± limn→∞ α n = 0. (2) |α| > 1 , k 1 α < 1, Kâ (1) (Ø, k limn→∞ 1 α n = 0 ¤± limn→∞ α n = ∞. 7/54