计算limx→无穷 (1+x)(1+x^2)(1+x^4)...(1+x^(2^n)),丨x丨<1
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计算limx→无穷 (1+x)(1+x^2)(1+x^4)...(1+x^(2^n)),丨x丨<1
应该是对|x| < 1,求lim{n → ∞} (1+x)(1+x^2)(1+x^4)...(1+x^(2^n))吧?
注意到(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^(2^n))
= (1-x^2)(1+x^2)(1+x^4)...(1+x^(2^n))
= (1-x^4)(1+x^4)...(1+x^(2^n))
= (1-x^8)...(1+x^(2^n))
...
= 1-x^(2^(n+1)).
所求极限可化为lim{n → ∞} (1-x^(2^(n+1)))/(1-x)
= (1-lim{n → ∞} x^(2^(n+1)))/(1-x)
= 1/(1-x) (|x| < 1故lim{n → ∞} x^(2^(n+1)) = 0).
注意到(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^(2^n))
= (1-x^2)(1+x^2)(1+x^4)...(1+x^(2^n))
= (1-x^4)(1+x^4)...(1+x^(2^n))
= (1-x^8)...(1+x^(2^n))
...
= 1-x^(2^(n+1)).
所求极限可化为lim{n → ∞} (1-x^(2^(n+1)))/(1-x)
= (1-lim{n → ∞} x^(2^(n+1)))/(1-x)
= 1/(1-x) (|x| < 1故lim{n → ∞} x^(2^(n+1)) = 0).
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