因为 1+1> an 综上,iman存在,令iman=A 又an1=√2an,故an=2an,因此lan=2lman,即2=2A 解得4=2,A2=0(舍去),故iman=2 (2)易知xn>0,先证此数列单调有界: 当n=1时,x=1≤2,当n>1时,x=1+—m,≤2,即xn≤2(m=1,2,…),即 此数列有界; 0,故 x -x )-(1 xn-1+1(xn+1xn-1+1) >0 (xn+1)xn-1+1)…(x2+1)x+1) 综上,imxn存在,令imxn=A 因此imxn=1+ lim x,+ 解得A= A (舍去),故lmxn 57…(2n-1),(2n)!!=2.4.6.8…(2n) 设 (2n-1)!! ,试证明 并求极限lm (2n)! 分<xn< 证易知 (2(n+1)-1)!2n+1 (2(n+1)!! 当n=1时 假设n=k时 则当 =k+1时4 因为 2 1 2 ( 2) 2 2 2 n n nn n n nn nn nn a a aa a a aa aa aa + − − − −= −= = + + , 由 2 n a < , 故 1 0 n n a a + − > , 即 n n 1 a a + > . 综上, lim n n a →∞ 存在, 令 lim n n a A →∞ = . 又 1 2 n n a a + = , 故 2 1 2 n n a a + = , 因此 2 1 lim 2 lim n n n n a a + →∞ →∞ = , 即 2 A = 2A , 解得 1A = 2 , 2 A = 0 (舍去), 故 lim 2 n n a →∞ = . (2) 易知 0 n x > , 先证此数列单调有界: 当 n =1时, 1x = ≤1 2 , 当 n >1时, 1 1 1 2 1 n n n x x x − − = + ≤ + , 即 2( 1,2, ) n x n ≤ = " , 即 此数列有界; 又 2 1 1 0 2 x x − => , 故 1 1 1 1 1 (1 ) (1 ) 1 1 ( 1)( 1) n n nn n n n n nn x x xx x x x x xx − − + − − − − =+ −+ = + + ++ 2 1 1 21 0 ( 1)( 1) ( 1)( 1) n n x x xx xx − − == > + + ++ " " , 即 n n 1 x x + > . 综上, lim n n x →∞ 存在, 令 lim n n x A →∞ = . 又 1 1 1 1 n n n x x x − − = + + , 因此 1 1 lim lim 1 lim 1 n n n n n n x x x − →∞ →∞ − →∞ = + + , 即 1 1 A A A = + + , 解得 1 1 5 2 A + = , 1 1 5 2 A − = (舍去), 故 1 5 lim 2 n n x →∞ + = . 5. 记(2 1)!! 1 3 5 7 (2 1) n n − =⋅⋅⋅ − " , (2 )!! 2 4 6 8 (2 ) n n = ⋅⋅⋅ " . 设 (2 1)!! (2 )!! n n x n − = , 试证明 1 1 4 21 n x n n < < + , 并求极限 lim n n x →∞ . 证 易知 1 (2( 1) 1)!! 2 1 (2( 1))!! 2 2 n n n n x x n n + +− + = =⋅ + + , 当 n =1 时, 1 1 11 4 21 2 < =< x + , 假设 n k = 时, 1 1 4 21 k x k k < < + , 则当 n k = +1时