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平方差 =(1-1/2)(1+1/2)(1-1/3)(1+1/3)……(1-1/n)(1+1/n) =(1/2)(3/2)(2/3)(4/3)……[(n-1)/n][(n+1)/n] 中間約分 =(1/2)[(n+1)/n] =(n+1)/2n =(1+1/n)/2 所以極限=1/2
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