Question

∑(-1)^n[(2n-1)!/(2n)! ]^p ,p>0是实数证明p>2是绝对收敛,p

Answer 1

证明:

令An={(2n-1)!/(2n)!},

则An+1/An=(2n+2)/(2n+2)根(1×3)

4=(3+5)/2>根(3×5)

......

2n=[(2n-1)+(2n+1)]/2>根[(2n-1)(2n+1)]

将上述不等式相乘,得

2*4*...*(2n)>根(1×3)*根(3×5)*...*根[(2n-1)(2n+1)]

--->0无穷)lim[(2n-1)!/(2n)!]=0.

单调减,un极限为0,则p大于0时,级数条件收敛,

∑[1根(2n+1)]^p当p=2时等价于1/(2n+1),是发散的,

当p