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I'm not sure but,

I guess (from my book) an asymptotic approximation of $(1-\frac{1}{\sqrt{n}})^n$ would be

$$(1-\frac{1}{\sqrt{n}})^n \space\space \to \space\space e^{-\sqrt{n}-\frac{1}{2} + O(n^{-1/2})}$$ as n increases.

Can you explain to me why?

(That's just my guessing, so you can correct it)

Maths lover
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David Lee
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    By the Maclaurin series of $\log(1+x)$, $$ \log \left( {1 - \frac{1}{{\sqrt n }}} \right)^n = n\log \left( {1 - \frac{1}{{\sqrt n }}} \right) = n\left( { - \frac{1}{{\sqrt n }} - \frac{1}{{2n}} - \frac{1}{{3n^{3/2} }} - \ldots } \right)=\ldots $$ Can you finish? – Gary Mar 15 '24 at 07:02
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    Wow. Thank you. It was kinda easy. I never thought it, though. Thank you.:) – David Lee Mar 15 '24 at 07:14

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