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Find the asymptotics of $n(\frac{n-1}{n})^n$.

I know $f(x)$~ $g(x) $ if $lim\frac{f(x)}{g(x)}=1$ but I am unsure as to how I found $g(x)$

I found a solution $\frac{2n-1}{2e}$ but I am unsure where that comes from

Any help would be appreciated

Thanks

MCCR
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2 Answers2

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Since

$$\left(\frac{n-1}n\right)^n=\left(1-\frac1n\right)^n\xrightarrow[n\to\infty]{}e^{-1}$$

I'd say the sequence you have is divergent to $\;\infty\;$ , and in fact

$$\frac{n\left(\frac{n-1}n\right)^n}{ne^{-1}}\xrightarrow[n\to\infty]{}1$$

Did
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DonAntonio
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For any real $a$, $\left(\dfrac{n+a}{n}\right)^n =\left(1+\dfrac{a}{n}\right)^n \approx e^a $, so $n\left(\dfrac{n+a}{n}\right)^n \approx ne^a $, and this diverges for all real $a$.

marty cohen
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