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

Why is this $\{(1+\frac1{a_n})^{a_n}=e\}$ true when: $a_n \to -\infty$

$a_n$ is a sequence.


share|cite|improve this question
Do you know that it holds if $a_n\longrightarrow +\infty$? – Git Gud Nov 27 '13 at 15:12
You mean $+ \infty$? – BIS HD Nov 27 '13 at 15:13
Yes I know about plus infinity, I ask about minus infinity, this is not a typo. It just doesn't make any sense... – GinKin Nov 27 '13 at 15:14
up vote 3 down vote accepted

Hint: Put $b_n=-a_n,$ and note then that $$\left(1+\frac1{a_n}\right)^{a_n}=\left(1+\frac{-1}{b_n}\right)^{-b_n}=\left(\left(1+\frac{-1}{b_n}\right)^{b_n}\right)^{-1}.$$ What is the limit of the expression inside the outermost parentheses as $n\to\infty$ (so that $b_n\to\infty$)?

share|cite|improve this answer
Nice work! I'll upvote at night. :) – Ahaan S. Rungta Nov 27 '13 at 15:18
I guess the minus 1 in the numerator doesn't interrupt the inside expression from going to e. But then it'll become $\frac1e$. I'm probably wrong. – GinKin Nov 27 '13 at 15:22
Actually, for any real $x,$ we have $$\lim_{t\to\infty}\left(1+\frac{x}{t}\right)^t=e^x,$$ as is discussed here, so in particular, when $x=-1,$ we see that...what? – Cameron Buie Nov 27 '13 at 16:08
Woah! Thanks!!! – GinKin Nov 27 '13 at 17:32
Thank you, @Ahaan! Glad you liked it. – Cameron Buie Nov 28 '13 at 5:13

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.