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

$a_n$ is a sequence.

Thanks.

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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
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1 Answer 1

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 $b_n\to\infty$)?

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Nice work! I'll upvote at night. :) –  Ahaan 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
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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
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