设zn=((1-i)/2)^n,n∈N,Sn=|z2-z1|+|z3-z2|+...+|Zn+1-Zn|,求极限Sn?
如题
人气：254 ℃ 时间：2020-03-24 13:35:42
解答
|Zn+1-Zn|=|((1-i)/2)^(n+1)-((1-i)/2)^n|=|((1-i)/2)^n|*|(1-i)/2-1|=|(1-i)/2|^n*1/√2=(1/√2)^(n+1)Sn=1/√2+(1/√2)^2+.(1/√2)^(n+1)等比数列极限Sn=a1/(1-q)极限Sn=1/√2/(1-1/√2)=√2+1
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