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Fabien
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I don't know if you are familiar with Taylor series, but if yes : $$\left(1-\frac1p\right)^n=\exp\left(n\ln\left(1-\frac1p\right)\right)=\exp\left(-n\left(\frac1p+o\left(\frac1p\right)\right)\right)$$ and this gives : $$\left(1-\frac1p\right)^n\sim \exp\left(-\frac{n}{p}\right)$$$$\left(1-\frac1p\right)^n\sim_{p>>1} \exp\left(-\frac{n}{p}\right)$$

Note than there's no assumption on n.

I don't know if you are familiar with Taylor series, but if yes : $$\left(1-\frac1p\right)^n=\exp\left(n\ln\left(1-\frac1p\right)\right)=\exp\left(-n\left(\frac1p+o\left(\frac1p\right)\right)\right)$$ and this gives : $$\left(1-\frac1p\right)^n\sim \exp\left(-\frac{n}{p}\right)$$

Note than there's no assumption on n.

I don't know if you are familiar with Taylor series, but if yes : $$\left(1-\frac1p\right)^n=\exp\left(n\ln\left(1-\frac1p\right)\right)=\exp\left(-n\left(\frac1p+o\left(\frac1p\right)\right)\right)$$ and this gives : $$\left(1-\frac1p\right)^n\sim_{p>>1} \exp\left(-\frac{n}{p}\right)$$

Note than there's no assumption on n.

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Fabien
  • 3k
  • 1
  • 16
  • 27

I don't know if you are familiar with Taylor series, but if yes : $$\left(1-\frac1p\right)^n=\exp\left(n\ln\left(1-\frac1p\right)\right)=\exp\left(-n\left(\frac1p+o\left(\frac1p\right)\right)\right)$$ and this gives : $$\left(1-\frac1p\right)^n\sim \exp\left(-\frac{n}{p}\right)$$

Note than there's no assumption on n.