lim(n→∞)(1+1/2)(1+1/2^2).......(1+1/2^2n)
=lim(n→∞)(1-1/2)(1+1/2)(1+1/2^2).......(1+1/2^2n)/(1-1/2)
=2lim(n→∞)(1-1/2^4n)
=2本回答被提问者采纳
lim(n→正无穷)(1+1\/2)(1+1\/2^2)(1+1\/2^4)...(1+1\/2^2n)=
lim(n→正无穷)（1+1\/2）(1+1\/2^2)(1+1\/2^4)...(1+1\/2^2n)lim(n→正无穷)2(1-1\/2^4n)=2
求极限lim(n→∝)(1+1\/2)(1+1\/2^2)(1+1\/2^4)…(1+1\/2^2n)
=2(1-1\/2)(1+1\/2)(1+1\/2^2)，，，(1+1\/2^2n)=2(1-1\/2^4n)则n→00，则A=2
求lim(n→∞) (1+1\/2)(1+1\/4)……(1+1\/2^n)
所以lim (1+1\/2)(1+1\/4)……(1+1\/2^n) （n→∞时)=2
lim(1+1\/2)(1+1\/4)(1+1\/16)……(1+1\/2^2n)
=lim2(1-1\/2^2n)(1+1\/2^2n)=lim2(1-1\/2^4n)=2 (当n--->∞时)
( 1+1\/2)(1+1\/2^2)(1+1\/2^4)……(1+\/2^2n)的极限n趋向无穷怎么算?)
( 1+1\/2)(1+1\/2^2)(1+1\/2^4)……(1+\/2^(2^n))的极限n趋向无穷=2(1-1\/2)(1+1\/2)(1+1\/2^2)(1+1\/2^4)……(1+\/2^(2^n))的极限n趋向无穷=2(1-1\/2^2)(1+1\/2^2)(1+1\/2^4)……(1+\/2^(2^n))的极限n趋向无穷=2(1-1\/2^4)(1+1\/2^...
求lim(1+1\/2+1\/4+……+1\/2n)的极限
xn=1+1\/2+1\/4+..+1\/2n =1+1\/2+1\/4+..+1\/n+(1\/2n+1\/2n)-1\/2n =1+1\/2+1\/2+..+1\/n+1\/n-1\/2n =...=2-1\/2n lim 2 =2 lim1\/2n=0 所以lim1+1\/2+1\/4+..+1\/2n=2
limn趋于无穷大(1+1\/2^1)(1+1\/2^2)...(1+1\/2^n)=?
回答：如有不明白,可以追问。如有帮助,记得采纳,谢谢
怎样证明不等式 (1+1\/1^2)(1+1\/2^2)(1+1\/3^2)……(1+1\/n^2)<e^2
不知道怎么搞的提示长度超了限制，私下联系我吧。令f(x)=(1+x)^2(1-x^2)^2n,在x=0处做泰勒展开得到f(x)=1+(2+2n)x+o(x),所以x≤1时有f(x)>1+x^2;令x=1\/(n+1)代入即得(1+1\/(n+1))^2(1-1\/(n+1)^2)^(2n)>1+1\/(n+1)^2 ...
你能求出(1+1\/2)(1+1\/4)(1+1\/16)...(1+1\/2的2n次方)的值吗?
解如图。
(1+1\/2)(1+1\/4)(1+1\/16)(1+1\/256)…(1+1\/2的2n次方)
原式=2*(1-1\/2)(1+1\/2)(1+1\/4)(1+1\/8)……(1+1\/2^2^n)=2*(1-1\/4)(1+1\/4)(1+1\/8)……(1+1\/2^2^n)=2*(1-1\/8)(1+1\/8)……(1+1\/2^2^n)反复用平方差=2*[1-1\/2^(2^(n+1))]=2-2\/2^(2^(n+1))]不懂请追问,满意请给好评,谢谢了...
