已知数列An满足a1=2 A(n+1)=2-1/An 求:令Tn=An/A(n+1)的求和,证明T

1个回答

  • (1)用归纳法证明An=(n+1)/n当n=1时A1=(1+1)/1=2显然成立.假设Ak=(k+1)/k,则有Ak+1=2-1/Ak=2-1/(k+1)/k=《(k+1)+1》/(k+1)即Ak+1也成立,则假设成立.即的证An=(n+1)/n (2)令Bn=An+1/An=n(n+2)/(n+1)²=1-1/(n+1)²>1-1/n(n+2)显然成立(因为(n+1)²>n(n+2) Bn>1-1/2《1/n-1/(n+2)》又因为Tn=A(n+1)/An的求和所以Tn>n-1/2《(1-1/3)+(1/2-1/4)+(1/3-1/5)+.+(1/n-1/(n+2))》=n-1/2《1+1/2-1/(n+1)-1/(n+2)》>n-1/2(1+1/2)=n-3/4即的证.《这个符号代替中挂号