Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
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
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 $-e$.

share|cite|improve this answer

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 $|x|<1$

share|cite|improve this answer
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 $\lim\limits_{n\rightarrow\infty}n\left[1-\left(1+\frac{1}{n}\right)^{e}\right]$

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.