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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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hint : $\ \bigl(1+\frac 1n\bigr)^e-1\sim\frac en\ $ for $n\gg 1$. –  Raymond Manzoni Oct 22 '12 at 15:43
    
@RaymondManzoni Thank you! –  Siyanda Oct 22 '12 at 15:47

4 Answers 4

up vote 0 down vote accepted

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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Please elaborate...I'm not entirely sure where this will lead me. This seems like a cool method –  Siyanda Oct 22 '12 at 15:54
    
@Dennis Gulko Obviously, $|x|<1$ :) –  M. Strochyk Oct 22 '12 at 15:59
    
@M. Strochyk: Sure ;) –  Dennis Gulko Oct 22 '12 at 16:08
    
@Siyanda: This just explains how Raymond Manzoni got that estimate. By setting $x=\frac1n$, $a=e$ and $n=2$. –  Dennis Gulko Oct 22 '12 at 16:10
    
@DennisGulko Ohhh! That explains it...thanks –  Siyanda Oct 22 '12 at 16:37

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 $$ -\sum_{k=1}^\infty\binom{e}{k}\frac{1}{n^{k-1}}. $$

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