Taylor expansion of function of exponential

Question

Show that if $n$ is large and $ne^{-t}\ll 1$, the expression $(1-e^{-t})^n$ reduces to $1-ne^{-t}$.

I know that the Taylor expansion of $e^{-t}$ when $t$ is small is $1-t$ but I am not sure how to use this in solving the problem. Also, why is the condition $ne^{-t}\ll 1$ necessary? Any hints are appreciated. Thank you.

Answer 1

By the binomial expansion,

$$(1-\epsilon)^n=1-n\epsilon+\frac{n(n-1)}2\epsilon^2-\frac{n(n-1)(n-2)}{3!}\epsilon^3\cdots \\=1-n\epsilon\left(1-\frac{n-1}2\epsilon+\frac{(n-1)(n-2)}{3!}\epsilon^2\cdots\right).$$

Then iff $n\epsilon\ll 1$,

$$(1-\epsilon)^n\approx1-n\epsilon.$$

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Working with $e^{-t}$ instead of $\epsilon$ is just "noise".

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Additional note:

If $n\epsilon\ll 1$, then

$$1-n\epsilon\approx 1$$ and the question is a little illogical. Better to ask about

$$(1-\epsilon)^n-1.$$

Answer 2

Use $$y^n=\sum_{i=0}^{n} {{n}\choose{i}} (y-1)^i.$$ With $y=1-e^{-t}$ you get $$(1-e^{-t})^n = \sum_{i=0}^{n} {{n}\choose{i}} (-e^{-t})^i\approx 1-ne^{-t}.$$