(1)當(dāng)q=1時(shí),Sn=n;S(n+1)=(n+1)
lim(Sn/Sn+1)=n/(n+1)=1
(2)當(dāng)q=-1時(shí),n為偶數(shù)Sn=0;S(n+1)=1,極限=0
n為奇數(shù),Sn=1;S(n+1)=0,極限不存在;
(3)當(dāng)q≠±1時(shí):
Sn=a1·(1-q^n)/(1-q)
S(n+1)=a1·[1-q^(n+1)]/(1-q)
Sn/S(n+1)=(1-q^n)/[1-q^(n+1)]
∴l(xiāng)im(Sn/Sn+1)=(1-q^n)/[1-q^(n+1)]
若|q|>1:
lim(Sn/Sn+1)=lim(1/q^n -1)/[1/q^n-q]=(-1)/(-q)=1/q
若|q|
已知無窮等比數(shù)列{an}首項(xiàng)為1,公比為q,前n項(xiàng)和為Sn,求lim(Sn/Sn+1)
已知無窮等比數(shù)列{an}首項(xiàng)為1,公比為q,前n項(xiàng)和為Sn,求lim(Sn/Sn+1)
數(shù)學(xué)人氣:420 ℃時(shí)間:2019-10-23 06:13:21
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