1+1/n+1/n^2+1/n^3+……=1/(1-1/n)怎么推导的?RT1+1/n+1/n^2+1/n^3+……=1/(1-1/n)推导过程

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1+1/n+1/n^2+1/n^3+……=1/(1-1/n)怎么推导的?RT1+1/n+1/n^2+1/n^3+……=1/(1-1/n)推导过程
1+1/n+1/n^2+1/n^3+……=1/(1-1/n)怎么推导的?
RT
1+1/n+1/n^2+1/n^3+……=1/(1-1/n)推导过程

1+1/n+1/n^2+1/n^3+……=1/(1-1/n)怎么推导的?RT1+1/n+1/n^2+1/n^3+……=1/(1-1/n)推导过程
等式左边是首项为1,公比为 1/n 的无穷等比数列的极限和,
利用等比数列求和公式,得左边= (k->∞) lim 1*{ 1 - (1/n)^k } / (1- 1/n) = 1/(1- 1/n)

等比数列求和
1+1/n+1/n^2+1/n^3+...-1/n(1+1/n+1/n^2+1/n^3+...)=1
1+1/n+1/n^2+1/n^3+...=1/(1-1/n)

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