数列求和、极限值P=1/4+2/8+3/16+4/32+.+n/[2的(n+1)次方]P=1/4+2/8+3/16+4/32+.+n/[2的(n+1)次方]求P的最简表达式及P的极限值.

求和： 头尾相加，乘以个数，除以2

P=1/4+2/8+3/16+4/32+.......+n/2^(n+1) (1)
1/2P= 1/8+2/16+3/32+......+(n-1)/2^(n+1)+n/2^(n+2) (2)
(1)-(2),得 1/2P=1/4+1/8+1/16+.....+1/2^(n+1)-n/2^(n+2)=1/4(1-1/2^n)/(1-1/2)-n/2^(n+2)=1/2-1/2^(n+1)-n/2^(n+2)
P=1-(n+2)/2^(n+1)
P的极致是1
注:lim n趋于正无穷(n+2)/2^(n+1)=lim n趋于正无穷1/((n+1)2^n)=0

典型的等差与等比乘积产生的混合数列，左右两边同时乘以或同时除以等比数列的公比，然后错位相减易可得P的表达式，求极限较易得。

P=1/4+2/8+3/16+4/32+.......+n/[2的(n+1)次方]
=1 /2^2 +2 / 2^3 +...+ n / 2^(n+1) (1)
1/2 *P =1 /2^3 +2 / 2^4 +...+ (n -1)/ 2^(n+1) +n / 2^(n+2) (2)
(1) - (2)得：1/2 *P= 1 / 2^2 +1 / 2^3 +....+ 1 / 2^(n+1) -n / 2^(n+2)
所以P=1 / 2 +1 / 2^2 +....+ 1 / 2^n -n / 2^(n+1)
=1 - 1 / (2^n) - n / 2^(n+2)

an = n/2^(n+1)
= (1/4)[n.(1/2)^(n-1) ]
consider
1+x+x^2+..+x^n= [x^(n+1) -1]/(x-1)
1+2x+3x^2+...+nx^(n-1)= ([x^(n+1) -1]/(x-1))'
= [nx^(n+1)-(n+1)x^n +1]/(x-1)^2
put x=1/2
∑(i:1->n) i.(1/2)^(i-1)
=4[n.(1/2)^(n+1)-(n+1)(1/2)^n +1]
=4[ 1- (n+2)(1/2)^(n+1) ]
P=1/4+2/8+3/16+4/32+.+n/[2^(n+1)]
=a1+a2+..+an
= (1/4)∑(i:1->n) i.(1/2)^(i-1)
= 1- (n+2)(1/2)^(n+1)
lim(n->∞)P = 1