Question

∑(1！+(2！)∧2+...+(n！)∧2)/(2n)！的敛散性

Answer 1

令Un=(1!+2!+……+n!)/(2n)!,U(n+1)=(1!+2!+……+n!+(n+1)!)/(2n+2)!,U(n+1)/Un=1/(2n+2)*(2n+1)+(n+1)!/(2n+2)*(2n+1)*(1!+2!+……+n!)/当n趋于无穷大时U(n+1)/Un趋于零（比式判别法的极限形式）即得证

Answer 2

S = 1.(1/2)^0+ 2.(1/2)^1+....+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1+ 2.(1/2)^2+....+n.(1/2)^n (2)
(1)-(2)
(1/2)S = (1+1/2+1/2^2+...+1/2^(n-1)) - n.(1/2)^n
= 2( 1- (1/2)^n ) - n.(1/2)^n
S =4( 1- (1/2)^n ) - 2n.(1/2)^n

an = (2n-1)/2^n
= n.(1/2)^(n-1) - 1/2^n
Sn = a1+a2+...+an
= S - (1- (1/2)^n )
= 3( 1- (1/2)^n ) - 2n.(1/2)^n
=3 -(2n+3).(1/2)^n

lim(n->∞) Sn =3