数列求和、极限值P=1/4+2/8+3/16+4/32+.+n/[2的(n+1)次方]

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