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$\lim_{n\rightarrow\infty}n\left[1-\left(1+\frac{1}{n}\right)^{e}\right]$

I tried playing around with the $\lim_{n\rightarrow\infty}n(1-\frac{1}{n})^n$ = $\frac{1}{e}$ identity but I can't really tell you where I'm headed with that one.

My gut keeps telling me the answer is infinity but my gut hasn't passed me an exam in years.

Some help would be nice.

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    @RaymondManzoni Thank you!2012-10-22

4 Answers 4

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Using Taylor expansions, $ n\left[1-\left(1+\frac1n\right)^e\right]=n\left[1-e^{e\,\log\left(1+\frac1n\right)}\right]=n\left[1-e^{e\,\left(\frac1n+o(1/n^2)\right)}\right]= n\left[1-\left(1+\frac{e}n+o(1/n^2)\right)\right]=-e+o(1/n). $

So the limit is $-e$.

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Expanding on @Raymond Manzoni's hint: if you know how to use Taylor expansion, it is useful to know that $(1+x)^a=1+ax+a(a-1)x^2+a(a-1)(a-2)x^3+...+a(a-1)...(a-n+2)x^{n-1}+o(x^n)$ for $|x|<1$

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    @DennisGulko Ohhh! That explains it...thanks2012-10-22
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You may use this $\lim\limits_{x \rightarrow{0}} \dfrac{(1+x)^\alpha-1}{x}=\alpha$, and put $x=\dfrac{1}{n}$ in $\lim\limits_{n\rightarrow\infty}n\left[1-\left(1+\frac{1}{n}\right)^{e}\right]$

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You can use binomial series to expand: $ \left(1+\frac{1}{n}\right)^e=\sum_{k=0}^\infty\binom{e}{k}\frac{1}{n^k}. $ Then $ 1-\left(1+\frac{1}{n}\right)^e=-\sum_{k=1}^\infty\binom{e}{k}\frac{1}{n^k}. $ Multiply by $n$ to get $ -\sum_{k=1}^\infty\binom{e}{k}\frac{1}{n^{k-1}}. $