lim(1+1/3+1/9+...+1/3^(n1))/(1+1/2+1/4+...+1/2^(n-1))
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![lim(1+1/3+1/9+...+1/3^(n1))/(1+1/2+1/4+...+1/2^(n-1))](/uploads/image/z/6980961-57-1.jpg?t=lim%281%2B1%2F3%2B1%2F9%2B...%2B1%2F3%5E%28n1%29%29%2F%281%2B1%2F2%2B1%2F4%2B...%2B1%2F2%5E%28n-1%29%29)
lim(1+1/3+1/9+...+1/3^(n1))/(1+1/2+1/4+...+1/2^(n-1))
lim(1+1/3+1/9+...+1/3^(n1))/(1+1/2+1/4+...+1/2^(n-1))
lim(1+1/3+1/9+...+1/3^(n1))/(1+1/2+1/4+...+1/2^(n-1))
首先对等比数列an=a1q^(n-1)
所有,他的和Sn=a1(1-q^n)/(1-q)
所以有1+1/3+1/9+……+1/3^(n-1)=1*(1-1/3^n)/(1-1/3)=3(1-1/3^n)/2=3/2-1/[2*3^(n-1)]
同理有1+1/2+1/4+……+1/2^(n-1)=1*(1-1/2^n)/(1-1/2)=2-1/2^(n-1)
所以有lim(1+1/3+1/9+...+1/3^(n1))/(1+1/2+1/4+...+1/2^(n-1))
=lim[3/2-1/2*3^(n-1)]/[2-1/2^)(n-1)]
当你趋于无穷大时有原式=3/2/2=3/4
首先对等比数列an=a1q^(n-1)
所有,他的和Sn=a1(1-q^n)/(1-q)
所以有1+1/3+1/9+……+1/3^(n-1)=1*(1-1/3^n)/(1-1/3)=3(1-1/3^n)/2=3/2-1/[2*3^(n-1)]
同理有1+1/2+1/4+……+1/2^(n-1)=1*(1-1/2^n)/(1-1/2)=2-1/2^(n-1)
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首先对等比数列an=a1q^(n-1)
所有,他的和Sn=a1(1-q^n)/(1-q)
所以有1+1/3+1/9+……+1/3^(n-1)=1*(1-1/3^n)/(1-1/3)=3(1-1/3^n)/2=3/2-1/[2*3^(n-1)]
同理有1+1/2+1/4+……+1/2^(n-1)=1*(1-1/2^n)/(1-1/2)=2-1/2^(n-1)
所以有lim(1+1/3+1/9+...+1/3^(n1))/(1+1/2+1/4+...+1/2^(n-1))
=lim[3/2-1/2*3^(n-1)]/[2-1/2^)(n-1)]
当你趋于无穷大时有原式=3/2/2=3/4
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